3  The fundamental theorem of calculus and the art of integration

3.1 The fundamental theorem

3.1.1 A step toward finding exact areas efficiently

Suppose we know the position function \(s(t)\) and the velocity function \(v(t)\) of an object moving in a straight line, and for the moment let us assume that \(v(t)\) is positive on \([a,b]\). Then, as shown in Figure 3.1, we know two different ways to compute the distance, \(D\), the object travels: one is that \(D = s(b) - s(a)\), the object’s change in position. The other is the area under the velocity curve, which is given by the definite integral, so \(D = \int_a^b v(t) \, dt\).

Figure 3.1: Finding distance traveled when we know a velocity function \(v\).

Since both of these expressions tell us the distance traveled, it follows that they are equal, so \[\begin{equation} s(b) - s(a) = \int_a^b v(t) \, dt\text{.} \end{equation} \tag{3.1}\] Equation 3.1 holds even when velocity is sometimes negative, because \(s(b) - s(a)\),the object’s change in position, is also measured by the net signed area on \([a,b]\) which is given by \(\int_a^b v(t) \, dt\).

Perhaps the most powerful fact Equation 3.1 reveals is that we can compute the integral’s value if we can find a formula for \(s\). Remember, \(s\) and \(v\) are related by the fact that \(v\) is the derivative of \(s\), or equivalently that \(s\) is an antiderivative of \(v\).

For example, suppose you are asked to determine the exact distance traveled on \([1,5]\) by an object with velocity function \(v(t) = 3t^2 + 40\) feet per second. The distance traveled on the interval \([1,5]\) is given by \[\begin{equation*} D = \int_1^5 v(t) \,dt = \int_1^5 (3t^2 + 40) \, dt = s(5) - s(1)\text{,} \end{equation*}\] where \(s\) is an antiderivative of \(v\). Now, the derivative of \(t^3\) is \(3t^2\) and the derivative of \(40t\) is \(40\), so it follows that \(s(t) = t^3 + 40t\) is an antiderivative of \(v\). Therefore, \[\begin{align*} D &= \int_1^5 3t^2 + 40 \, dt = s(5) - s(1)\\ &= (5^3 + 40 \cdot 5) - (1^3 + 40\cdot 1) = 284 \ \text{feet}\text{.} \end{align*}\]

Note the key lesson of the previous examples: to find the distance traveled, we need to compute the area under a curve, which is given by the definite integral. But to evaluate the integral, we can find an antiderivative, \(s\), of the velocity function, and then compute the total change in \(s\) on the interval. In particular, we can evaluate the integral without computing the limit of a Riemann sum.

Figure 3.2: The exact area of the region enclosed by \(v(t) = 3t^2 + 40\) on \([1,5]\).

It will be convenient to have a shorthand symbol for a function’s antiderivative. For a continuous function \(f\), we will often denote an antiderivative of \(f\) by \(F\), so that \(F'(x) = f(x)\) for all relevant \(x\). Using the notation \(V\) in place of \(s\) (so that \(V\) is an antiderivative of \(v\)) in Equation 3.1), we can write \[\begin{equation} V(b) - V(a) = \int_a^b v(t) \, dt\text{.} \end{equation} \tag{3.2}\]

Now, to evaluate the definite integral \(\int_a^b f(x) \, dx\) for an arbitrary continuous function \(f\), we could certainly think of \(f\) as representing the velocity of some moving object, and \(x\) as the variable that represents time. But Equation 3.1 and Equation 3.2 hold for any continuous velocity function, even when \(v\) is sometimes negative. So Equation 3.2 offers a shortcut route to evaluating any definite integral, provided that we can find an antiderivative of the integrand. The Fundamental Theorem of Calculus (FTC) summarizes these observations.

Theorem 3.1 (The Fundamental Theorem of Calculus) If \(f\) is a continuous function on \([a,b]\), and \(F\) is any antiderivative of \(f\), then \(\int_a^b f(x) \, dx = F(b) - F(a)\).

A common alternate notation for \(F(b) - F(a)\) is \[\begin{equation*} F(b) - F(a) = \left. F(x) \right|_a^b\text{,} \end{equation*}\] where we read the right hand side as “the function \(F\) evaluated from \(a\) to \(b\).” In this notation, the FTC says that \[\begin{equation*} \int_a^b f(x) \, dx = \left. F(x) \right|_a^b\text{.} \end{equation*}\]

The FTC opens the door to evaluating a wide range of integrals if we can find an antiderivative \(F\) for the integrand \(f\). For instance since \(\frac{d}{dx}[\frac{1}{3}x^3] = x^2\), the FTC tells us that \[\begin{align*} \int_0^1 x^2 \, dx =\mathstrut & \left. \frac{1}{3} \, x^3 \right|_0^1\\ =\mathstrut & \frac{1}{3} \, (1)^3 - \frac{1}{3} \, (0)^3\\ =\mathstrut & \frac{1}{3}\text{.} \end{align*}\]

But finding an antiderivative can be far from simple; it is often difficult or even impossible. While we can differentiate just about any function, even some relatively simple functions don’t have an elementary antiderivative.

Activity

Use the FTC to evaluate each of the following integrals exactly. For each, sketch a graph of the integrand on the relevant interval and write one sentence that explains the meaning of the value of the integral in terms of the (net signed) area bounded by the curve.

  1. \(\displaystyle \int_{-1}^4 (2-2x) \, dx\)
  2. \(\displaystyle \int_0^1 e^x \, dx\)
  3. \(\displaystyle \int_{-1}^{1} x^5 \, dx\)
  4. \(\displaystyle \int_0^2 (3x^3 - 2x^2 - e^x) \, dx\)

3.1.2 Basic antiderivatives

The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of \(g(x) = x^2\) is \(G(x) = \frac{1}{3} x^3\), combinations of \(f\) and \(g\) can be far more complicated.

What is involved in trying to find an antiderivative for each? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.

We do note that whenever we know the derivative of a function, we have a function-derivative pair, so we also know an antiderivative of a function. Clearly, every basic derivative rule leads us to such a pair, and thus to a known antiderivative.

In the next activity, you will construct a list of the basic antiderivatives. Those rules will help us antidifferentiate sums and constant multiples of basic functions. Before proceeding to build a list of common functions whose antiderivatives we know, we recall that each function has more than one antiderivative. Because the derivative of any constant is zero, we may add a constant of our choice to any antiderivative. For instance, we know that \(G(x) = \frac{1}{3}x^3\) is an antiderivative of \(g(x) = x^2\). But we could also have chosen \(G(x) = \frac{1}{3}x^3 + 7\), since in this case as well, \(G'(x) = x^2\). If \(g(x) = x^2\), we say that the general antiderivative of \(g\) is \[\begin{equation*} G(x) = \frac{1}{3}x^3 + C\text{,} \end{equation*}\] where \(C\) represents an arbitrary real number constant. Regardless of the formula for \(g\), including \(+C\) in the formula for its antiderivative \(G\) results in the most general possible antiderivative.

Our current interest in antiderivatives is so that we can evaluate definite integrals by the Fundamental Theorem of Calculus. For that task, the constant \(C\) is irrelevant, and we usually omit it. To see why, consider the definite integral \[\begin{equation*} \int_0^1 x^2 \, dx\text{.} \end{equation*}\]

For the integrand \(g(x) = x^2\), suppose we find and use the general antiderivative \(G(x) = \frac{1}{3} x^3 + C\). Then, by the Fundamental Theorem of Calculus, \[\begin{align*} \int_0^1 x^2 \, dx &= \left. \frac{1}{3} x^3 + C \right|_0^1\\ &= \left(\frac{1}{3} (1)^3 + C \right) - \left(\frac{1}{3} (0)^3 + C \right)\\ &= \frac{1}{3} + C - 0 - C\\ &= \frac{1}{3}\text{.} \end{align*}\]

Observe that the \(C\)-values appear as opposites in the evaluation of the integral and thus do not affect the definite integral’s value.

In the following activity, we work to build a list of basic functions whose antiderivatives we already know.

Activity

Use your knowledge of derivatives of basic functions to complete Table 3.1 of antiderivatives. For each entry, your task is to find a function \(F\) whose derivative is the given function \(f\). When finished, use Theorem 3.1 and the results in the table to evaluate the three given definite integrals.

Table 3.1: Familiar basic functions and their antiderivatives.
given function, \(f(x)\) antiderivative, \(F(x)\)
\(k\), (\(k\) is constant)
\(x^n\), \(n \neq -1\)
\(\frac{1}{x}\), \(x > 0\)
\(e^x\)
\(a^x\) \((a > 1)\)
  1. \(\displaystyle \displaystyle \int_0^1 \left(x^3 - x - e^x + 2\right) \,dx\)
  2. \(\displaystyle \displaystyle \int_0^1 (\sqrt{x}-x^2) \, dx\)

3.1.3 Exercises

  1. The velocity function is \(v(t) = - t^2 + 4 t - 3\) for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval \([-1,5]\).
  2. Evaluate the following integrals:
    1. \(\displaystyle \int_{2}^{4} \frac{1}{ x^2 } dx\)
    2. \(\displaystyle\int_{ 2 } ^ { 9 } (4 x + 10)\, dx\)
    3. \(\displaystyle\int_{ -6 } ^ { 6 } (36 -x^2)\, dx\)
    4. \(\displaystyle\int_{ -6 } ^ { 6 } (6 -x)(6+x)\, dx\)
    5. \(\displaystyle\int_{ 3 } ^ { 8 } \frac {8 x^2 + 3 } { \sqrt x }\,dx\)
  3. The instantaneous velocity (in meters per minute) of a moving object is given by the function \(v\) as pictured in Figure 3.3. Assume that on the interval \(0 \le t \le 4\), \(v(t)\) is given by \(v(t) = -\frac{1}{4}t^3 + \frac{3}{2}t^2 + 1\), and that on every other interval \(v\) is piecewise linear, as shown.
Figure 3.3: The velocity function of a moving body.
  1. Determine the exact distance traveled by the object on the time interval \(0 \le t \le 4\).
  2. Suppose that the velocity of the object is increased by a constant value \(c\) for all values of \(t\). What value of \(c\) will make the object’s total distance traveled on \([12,24]\) be 210 meters?
  1. We noted that for \(x \gt 0\), an antiderivative of \(f(x) = \frac{1}{x}\) is \(F(x) = \ln(x)\). Here we observe that a key difference between \(f(x)\) and \(F(x)\) is that \(f\) is defined for all \(x \ne 0\), while \(F\) is only defined for \(x \gt 0\), and see how we can actually define an antiderivative of \(f\) for all values of \(x\).
  1. Suppose that \(x \lt 0\), and let \(G(x) = \ln(-x)\). Compute \(G'(x)\).
  2. Explain why \(G\) is an antiderivative of \(f\) for \(x \lt 0\).
  3. Let \(H(x) = \ln(|x|)\), and recall that \[\begin{equation*} |x| = \begin{cases} -x, & \ \text{if} \ x \lt 0 \\ x, & \ \text{if} \ x \ge 0 \end{cases}\text{.} \end{equation*}\] Explain why \(H(x) = G(x)\) for \(x \lt 0\) and \(H(x) = F(x)\) for \(x \gt 0\).
  4. Now discuss why we say that the antiderivative of \(\frac{1}{x}\) is \(\ln(|x|)\) for all \(x \ne 0\).

3.2 Integration by substitution

In this section, we answer the following questions:

  1. How can we begin to find algebraic formulas for antiderivatives of more complicated algebraic functions?
  2. What is an indefinite integral and how is its notation used in discussing antiderivatives?
  3. How does the technique of \(u\)-substitution work to help us evaluate certain indefinite integrals, and how does this process rely on identifying function-derivative pairs?

Previously, we learned the key role that antiderivatives play in the process of evaluating definite integrals exactly. The Fundamental Theorem of Calculus tells us that if \(F\) is any antiderivative of \(f\), then

\[\begin{equation*} \int_a^b f(x) \, dx = F(b) - F(a)\text{.} \end{equation*}\]

Furthermore, we realized that each elementary derivative rule leads to a corresponding elementary antiderivative. Thus, if we wish to evaluate an integral such as \[\begin{equation*} \int_0^1 \left(x^3 - \sqrt{x} + 5^x \right) \,dx\text{,} \end{equation*}\] it is straightforward to do so, since we can easily antidifferentiate \(f(x) = x^3 - \sqrt{x} + 5^x\). Because one antiderivative of \(f\) is \(F(x) = \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\), the Fundamental Theorem of Calculus tells us that \[\begin{align*} \int_0^1 \left(x^3 - \sqrt{x} + 5^x\right) \,dx &= \left. \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\right|_0^1\\ &= \left( \frac{1}{4}(1)^4 - \frac{2}{3}(1)^{3/2} + \frac{1}{\ln(5)}5^1 \right) - \left( 0 - 0 + \frac{1}{\ln(5)}5^0 \right)\\ &= -\frac{5}{12} + \frac{4}{\ln(5)}\text{.} \end{align*}\]

Preview Activity

You have learned the Chain Rule before and how it can be applied to find the derivative of a composite function. In particular, if \(u\) is a differentiable function of \(x\), and \(f\) is a differentiable function of \(u(x)\), then \[\begin{equation*} \frac{d}{dx} \left[ f(u(x)) \right] = f'(u(x)) \cdot u'(x)\text{.} \end{equation*}\]

In words, we say that the derivative of a composite function \(c(x) = f(u(x))\), where \(f\) is considered the “outer” function and \(u\) the “inner” function, is “the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.”

  1. For each of the following functions, use the Chain Rule to find the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of \(g(x)\) should be labeled \(g'(x)\)).
    1. \(\displaystyle g(x) = e^{3x}\)
    2. \(\displaystyle q(x) = (2-7x)^4\)
    3. \(\displaystyle r(x) = 3^{4-11x}\)
  2. For each of the following functions, use your work in Item 1 to help you determine the general antiderivative1 of the function. Label each antiderivative by name (e.g., the antiderivative of \(m\) should be called \(M\)). In addition, check your work by computing the derivative of each proposed antiderivative.
    1. \(\displaystyle m(x) = e^{3x}\)
    2. \(\displaystyle v(x) = (2-7x)^3\)
    3. \(\displaystyle w(x) = 3^{4-11x}\)
  3. Based on your experience in Items 1 and 2, conjecture an antiderivative for each of the following functions. Test your conjectures by computing the derivative of each proposed antiderivative.
    1. \(\displaystyle b(x) = (4x+7)^{11}\)
    2. \(\displaystyle c(x) = xe^{x^2}\)

3.2.1 Reversing the Chain Rule: First Steps

Whenever \(f\) is a familiar function whose antiderivative is known and \(u(x)\) is a linear function, it is straightforward to antidifferentiate a function of the form \[\begin{equation*} h(x) = f(u(x))\text{.} \end{equation*}\]

For example, determine the general antiderivative of \[\begin{equation*} h(x) = (5x-3)^6\text{.} \end{equation*}\] Check the result by differentiating.

For this composite function, the outer function \(f\) is \(f(u) = u^6\), while the inner function is \(u(x) = 5x - 3\). Since the antiderivative of \(f\) is \(F(u) = \frac{1}{7}u^7+C\), we see that the antiderivative of \(h\) is \[\begin{equation*} H(x) = \frac{1}{7} (5x-3)^7 \cdot \frac{1}{5} + C = \frac{1}{35} (5x-3)^7 + C\text{.} \end{equation*}\]

The inclusion of the constant \(\frac{1}{5}\) is essential precisely because the derivative of the inner function is \(u'(x) = 5\). Indeed, if we now compute \(H'(x)\), we find by the Chain Rule (and Constant Multiple Rule) that \[\begin{equation*} H'(x) = \frac{1}{35} \cdot 7(5x-3)^6 \cdot 5 = (5x-3)^6 = h(x)\text{,} \end{equation*}\] and thus \(H\) is indeed the general antiderivative of \(h\).

Hence, in the special case where the outer function is familiar and the inner function is linear, we can antidifferentiate composite functions according to the following rule.

Theorem 3.2 If \(h(x) = f(ax + b)\) and \(F\) is a known algebraic antiderivative of \(f\), then the general antiderivative of \(h\) is given by \[\begin{equation*} H(x) = \frac{1}{a} F(ax+b) + C\text{.} \end{equation*}\]

It is useful to have shorthand notation that indicates the instruction to find an antiderivative. Thus, in a similar way to how the notation \[\begin{equation*} \frac{d}{dx} \left[ f(x) \right] \end{equation*}\] represents the derivative of \(f(x)\) with respect to \(x\), we use the notation of the indefinite integral, \[\begin{equation*} \int f(x) \, dx \end{equation*}\] to represent the general antiderivative of \(f\) with respect to \(x\). Returning to the earlier example with \(h(x) = (5x-3)^6\), we can rephrase the relationship between \(h\) and its antiderivative \(H\) through the notation \[\begin{equation*} \int (5x-3)^6 \, dx = \frac{1}{35} (5x-6)^7 + C\text{.} \end{equation*}\]

When we find an antiderivative, we will often say that we evaluate an indefinite integral. Just as the notation \(\frac{d}{dx} [ \square ]\) means “find the derivative with respect to \(x\) of \(\square\),” the notation \(\int \square \, dx\) means “find a function of \(x\) whose derivative is \(\square\).”

Activity

Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.

  1. \(\displaystyle \int \frac{1}{11x - 9} \, dx\)
  2. \(\displaystyle \int 5^{-x}\, dx\)

3.2.2 Reversing the Chain Rule: \(u\)-substitution

A natural question arises from our recent work: what happens when the inner function is not linear? For example, can we find antiderivatives of such functions as \[\begin{equation*} g(x) = x e^{x^2} \ \text{and} \ h(x) = e^{x^2}? \end{equation*}\]

It is important to remember that differentiation and antidifferentiation are almost inverse processes (that they are not is due to the \(+C\) that arises when antidifferentiating). This almost-inverse relationship enables us to take any known derivative rule and rewrite it as a corresponding rule for an indefinite integral. For example, since \[\begin{equation*} \frac{d}{dx} \left[x^5\right] = 5x^4\text{,} \end{equation*}\] we can equivalently write \[\begin{equation*} \int 5x^4 \, dx = x^5 + C\text{.} \end{equation*}\]

Recall that the Chain Rule states that \[\begin{equation*} \frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x)\text{.} \end{equation*}\]

Restating this relationship in terms of an indefinite integral, \[\begin{equation} \int f'(g(x)) g'(x) \, dx = f(g(x))+C\text{.} \end{equation} \tag{3.3}\]

Equation 3.3 tells us that if we can view a given function as \(f'(g(x)) g'(x)\) for some appropriate choices of \(f\) and \(g\), then we can antidifferentiate the function by reversing the Chain Rule. Note that both \(g(x)\) and \(g'(x)\) appear in the form of \(f'(g(x)) g'(x)\); we will sometimes say that we seek to identify a function-derivative pair (\(g(x)\) and \(g'(x)\)) when trying to apply the rule in Equation 3.3.

If we can identify a function-derivative pair, we will introduce a new variable \(u\) to represent the function \(g(x)\). With \(u = g(x)\), it follows in Leibniz notation that \(\frac{du}{dx} = g'(x)\), so that in terms of differentials, \(du = g'(x)\, dx\). Now converting the indefinite integral to a new one in terms of \(u\), we have \[\begin{equation*} \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du\text{.} \end{equation*}\]

Provided that \(f'\) is an elementary function whose antiderivative is known, we can easily evaluate the indefinite integral in \(u\), and then go on to determine the desired overall antiderivative of \(f'(g(x)) g'(x)\). We call this process \(u\)-substitution, and summarize the rule as follows:

With the substitution \(u = g(x)\), \[\begin{equation*} \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du = f(u) + C = f(g(x)) + C\text{.} \end{equation*}\]

To see \(u\)-substitution at work, we consider evaluating the indefinite integral \[\begin{equation*} \int x^3 \cdot (7x^4 + 3)^2 \, dx \end{equation*}\] and check the result by differentiating.

We can make two algebraic observations regarding the integrand, \(x^3 \cdot (7x^4 + 3)^2\). First, \((7x^4 + 3)^2\) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating. Second, \(x^3\) is almost the derivative of \((7x^4 + 3)\); the only issue is a missing constant. Thus, \(x^3\) and \((7x^4 + 3)\) are nearly a function-derivative pair. Furthermore, we know the antiderivative of \(f(u) = u^2\). The combination of these observations suggests that we can evaluate the given indefinite integral by reversing the chain rule through \(u\)-substitution.

Letting \(u\) represent the inner function of the composite function \((7x^4 + 3)^2\), we have \(u = 7x^4 + 3\), and thus \(\frac{du}{dx} = 28x^3\). In differential notation, it follows that \(du = 28x^3 \, dx\), and thus \(x^3 \, dx = \frac{1}{28} \, du\). The original indefinite integral may be slightly rewritten as \[\begin{equation*} \int (7x^4 + 3)^2 \cdot x^3 \, dx\text{,} \end{equation*}\] and so by substituting \(u\) for \(7x^4 + 3\) and \(\frac{1}{28} \, du\) for \(x^3 \, dx\), it follows that \[\begin{equation*} \int (7x^4 + 3)^2 \cdot x^3 \, dx = \int u^2 \cdot \frac{1}{28} \, du\text{.} \end{equation*}\]

Now we may evaluate the easier integral in \(u\), and then replace \(u\) by the expression \(7x^4 + 3\). Doing so, we find \[\begin{align*} \int (7x^4 + 3)^2 \cdot x^3 \, dx &= \int u^2 \cdot \frac{1}{28} \, du\\ &= \frac{1}{28} \int u^2 \, du\\ &= \frac{1}{28} (\frac{u^3}{3}) + C\\ &= \frac{1}{28} \frac{(7x^4 + 3)^3}{3} + C\text{.} \end{align*}\]

To check our work, we observe by the Chain Rule that \[\begin{equation*} \frac{d}{dx} \left[\frac{1}{28}\frac{(7x^4 + 3)^3}{3}\right] = \frac{1}{28} \cdot\frac{1}{3} \cdot 3 (7x^4 + 3)^2 \cdot 28x^3 = (7x^4 + 3)^2 \cdot x^3\text{,} \end{equation*}\] which is indeed the original integrand.

The \(u\)-substitution worked because the function multiplying \(\sin (7x^4 + 3)\) was \(x^3\). If instead that function was \(x^2\) or \(x^4\), the substitution process would not have worked. This is one of the primary challenges of antidifferentiation: slight changes in the integrand make tremendous differences. For instance, we can use \(u\)-substitution with \(u = x^2\) and \(du = 2xdx\) to find that \[\begin{align*} \int xe^{x^2} \, dx &= \int e^u \cdot \frac{1}{2} \, du\\ &= \frac{1}{2} \int e^u \, du\\ &= \frac{1}{2} e^u + C\\ &= \frac{1}{2} e^{x^2} + C\text{.} \end{align*}\]

However, for the similar indefinite integral \[\begin{equation*} \int e^{x^2} \, dx\text{,} \end{equation*}\] the \(u\)-substitution \(u = x^2\) is no longer possible because the factor of \(x\) is missing. Hence, part of the lesson of \(u\)-substitution is just how specialized the process is: it only applies to situations where, up to a missing constant, the integrand is the result of applying the Chain Rule to a different, related function.

Activity

Evaluate each of the following indefinite integrals by using these steps:

  1. Find two functions within the integrand that form (up to a possible missing constant) a function-derivative pair.
  2. Make a substitution and convert the integral to one involving \(u\) and \(du\).
  3. Evaluate the new integral in \(u\).
  4. Convert the resulting function of \(u\) back to a function of \(x\) by using your earlier substitution.
  5. Check your work by differentiating the function of \(x\). You should come up with the integrand originally given.
  1. \(\displaystyle \int \frac{x^2}{5x^3+1} \, dx\)
  2. \(\displaystyle \int t^{3}\left(t^{4}-4\right)^{2} \,dt\)
  3. \(\displaystyle \int \frac{\ln^{8}\left(z\right)}{z}\,dz\)
  4. \(\displaystyle \int \frac{e^{5x}}{1+e^{5x}} \,dx\)
  5. \(\displaystyle \int \frac{2e^{4\sqrt{y}}}{\sqrt{y}} \,dy\)

3.2.3 Evaluating Definite Integrals via \(u\)-substitution

We have introduced \(u\)-substitution as a means to evaluate indefinite integrals of functions that can be written, up to a constant multiple, in the form \(f(g(x))g'(x)\). This same technique can be used to evaluate definite integrals involving such functions, though we need to be careful with the corresponding limits of integration. Consider, for instance, the definite integral \[\begin{equation*} \int_2^5 xe^{x^2} \, dx\text{.} \end{equation*}\]

Whenever we write a definite integral, it is implicit that the limits of integration correspond to the variable of integration. To be more explicit, observe that \[\begin{equation*} \int_2^5 xe^{x^2} \, dx = \int_{x=2}^{x=5} xe^{x^2} \, dx\text{.} \end{equation*}\]

When we execute a \(u\)-substitution, we change the variable of integration; it is essential to note that this also changes the limits of integration. For instance, with the substitution \(u = x^2\) and \(du = 2x \, dx\), it also follows that when \(x = 2\), \(u = 2^2 = 4\), and when \(x = 5\), \(u = 5^2 = 25\). Thus, under the change of variables of \(u\)-substitution, we now have \[\begin{align*} \int_{x=2}^{x=5} xe^{x^2} \, dx &= \int_{u=4}^{u=25} e^{u} \cdot \frac{1}{2} \, du\\ &= \left. \frac{1}{2}e^u \right|_{u=4}^{u=25}\\ &= \frac{1}{2}e^{25} - \frac{1}{2}e^4\text{.} \end{align*}\] Alternatively, we could consider the related indefinite integral \(\int xe^{x^2} \, dx\), find the antiderivative \(\frac{1}{2}e^{x^2}\) through \(u\)-substitution, and then evaluate the original definite integral. With that method, we’d have \[\begin{align*} \int_{2}^{5} xe^{x^2} \, dx &= \left. \frac{1}{2}e^{x^2} \right|_{2}^{5}\\ &= \frac{1}{2}e^{25} - \frac{1}{2}e^4\text{,} \end{align*}\] which is, of course, the same result.

Activity

Evaluate each of the following definite integrals exactly through an appropriate \(u\)-substitution.

  1. \(\displaystyle \int_1^2 \frac{x}{1 + 4x^2} \, dx\)
  2. \(\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx\)

3.2.4 Exercises

  1. Evaluate the following integrals and check your answers.
    1. \(\displaystyle\int x^2\left(x^3+1\right)\, dx\)
    2. \(\displaystyle\int \frac{e^x}{e^x+1}\,dx\)
    3. \(\displaystyle\int \frac{x}{x^2-4x+4}\,dx\)
    4. \(\displaystyle\int e^{e^x} e^x \, dx\)
  2. Consider the indefinite integral \(\int x \sqrt{x-1} \, dx\). At first glance, this integrand may not seem suited to substitution due to the presence of \(x\) in separate locations in the integrand. Nonetheless, using the composite function \(\sqrt{x-1}\) as a guide, let \(u = x-1\).
    1. Determine expressions for both \(x\) and \(dx\) in terms of \(u\).
    2. Convert the given integral in \(x\) to a new integral in \(u\).
    3. Evaluate the integral in (b) by noting that \(\sqrt{u} = u^{1/2}\) and observing that it is now possible to rewrite the integrand in \(u\) by expanding through multiplication.
    4. Evaluate each of the integrals \(\int x^2 \sqrt{x-1} \, dx\) and \(\int x \sqrt{x^2 - 1} \, dx\).
    5. Write a paragraph to discuss the similarities among the three indefinite integrals in this problem and the role of substitution and algebraic rearrangement in each.

3.3 Integration by parts

In this section, we answer the following questions:

  1. How do we evaluate indefinite integrals that involve products of basic functions such as \(\int x e^x \, dx\)?
  2. What is the method of integration by parts and how can we consistently apply it to integrate products of basic functions?
  3. How does the algebraic structure of functions guide us in identifying \(u\) and \(dv\) in using integration by parts?

In Section 3.2, we learned the technique of \(u\)-substitution for evaluating indefinite integrals. For example, the indefinite integral \(\int x^3 (7x^4+3)^2 \, dx\) is perfectly suited to \(u\)-substitution, because one factor is a composite function and the other factor is the derivative (up to a constant) of the inner function. Recognizing the algebraic structure of a function can help us to find its antiderivative. Next we consider integrands with a different elementary algebraic structure: a product of basic functions.

Preview Activity

To that end, we refresh our understanding of the Product Rule and then investigate some indefinite integrals that involve products of basic functions.

Recall that the Product Rule was employed to differentiate a product of two functions. In particular, recall that if \(f\) and \(g\) are differentiable functions of \(x\), then \[\begin{equation*} \frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + g(x) \cdot f'(x)\text{.} \end{equation*}\]

  1. For each of the following functions, use the Product Rule to find the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of \(g(x)\) should be labeled \(g'(x)\)).
    1. \(\displaystyle h(x) = xe^x\)
    2. \(\displaystyle p(x) = x\ln(x)\)
  2. Use your work in Item 1 to help you evaluate the following indefinite integrals. Use differentiation to check your work.
    1. \(\displaystyle \int \left(xe^x + e^x\right) \, dx\)
    2. \(\displaystyle \int \left(1 + \ln(x)\right) \, dx\)
  3. Observe that the examples in Item 2 work nicely because of the derivatives you were asked to calculate in Item 1. Each integrand in Item 2 is precisely the result of differentiating one of the products of basic functions found in Item 1. To see a slightly different structure, consider how to evaluate \[\begin{equation*} \int \left(\ln(x)\right)^2 \, dx\text{.} \end{equation*}\]
    1. First, observe that \[\begin{equation*} \frac{d}{dx} \left[ x\left(\ln(x)\right)^2 \right] = 2\ln(x) + \left(\ln(x)\right)^2\text{.} \end{equation*}\] Integrating both sides and using the fact that the integral of a sum is the sum of the integrals, we find that \[\begin{equation*} \int \left(\frac{d}{dx} \left[ x\left(\ln(x)\right)^2 \right] \right) \, dx = \int 2\ln(x) \, dx + \int \left(\ln(x)\right)^2 \, dx\text{.} \end{equation*}\] In this last equation, evaluate the indefinite integral on the left side as well as the rightmost indefinite integral on the right.
    2. Solve this last equation for the expression \(\int\left(\ln(x)\right)^2 \, dx\) and simplify using calculations already made in Item 2b.

3.3.1 Reversing the Product Rule: Integration by Parts

The previous Preview Activity provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that \[\begin{equation*} \frac{d}{dx} \left[ f(x) g(x) \right] = f(x) g'(x) + g(x) f'(x)\text{.} \end{equation*}\]

Integrating both sides of this equation indefinitely with respect to \(x\), we find \[\begin{equation} \int \frac{d}{dx} \left[ f(x) g(x) \right] \, dx = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.} \end{equation} \tag{3.4}\]

On the left side of Equation 3.4, we have the indefinite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have \[\begin{equation} f(x) g(x) = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.}\label{E-intprod2} \end{equation}\]

We solve for the first indefinite integral on the left to generate the rule \[\begin{equation} \int f(x) g'(x) \, dx = f(x) g(x) - \int g(x) f'(x) \, dx\text{.} \end{equation} \tag{3.5}\]

Often we express Equation 3.5 in terms of the variables \(u\) and \(v\), where \(u = f(x)\) and \(v = g(x)\). In differential notation, \(du = f'(x) \, dx\) and \(dv = g'(x) \, dx\), so we can state the rule for Integration by Parts in its most common form as follows:

\[\begin{equation*} \int u \, dv = uv - \int v \, du\text{.} \end{equation*}\]

To apply integration by parts, we look for a product of basic functions that we can identify as \(u\) and \(dv\). If we can antidifferentiate \(dv\) to find \(v\), and evaluating \(\int v \, du\) is not more difficult than evaluating \(\int u \, dv\), then this substitution usually proves to be fruitful.

To demonstrate, let us consider \[\int te^{-t}\,dt.\] Some choices for \(u\) and \(dv\) are

  1. \(u=t\) and \(dv=e^{-t}\, dt\)
  2. \(u=e^{-t}\) and \(dv=t\,dt\)

The second option gives \(du=-e^{-t}\,dt\) and \(v=t^2/2\). We then have \[\begin{eqnarray}\int te^{-t}\,dt &=& \frac{t^3}{2}-\int \frac{t^2}{2}\left(-e^{-t}\right)\,dt \\ &=& \frac{t^3}{2}+\frac{1}{2}\int t^2e^{-t}\,dt. \end{eqnarray}\] Observe that the problem became even more complicated. Instead of evaluating our original integral \(\int te^{-t}\,dt\), we are faced with evaluating \(\int t^2e^{-t}\,dt\). Therefore, this setup may not be the most ideal way to proceed with integration by parts.

You should now try using the first option where \(u=t\) and \(dv=e^{-t}\, dt\). Determine whether integration by parts is more feasible.

3.3.2 Some Subtleties with Integration by Parts

Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating \[\begin{equation*} \int \ln(x) \, dx\text{.} \end{equation*}\]

Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that \(\ln(x) = \ln(x) \cdot 1\), and realize that we know the derivative of \(\ln(x)\) as well as an antiderivative of \(1\), we see the possibility for the substitution \(u = \ln(x)\) and \(dv = 1 \, dx\).

In a related problem, consider \(\int t^3 e^{t^2} \, dt\). Observe that there is a composite function present in \(e^{t^2}\), but there is not an obvious function-derivative pair, as we have \(t^3\) (rather than simply \(t\)) multiplying \(e^{t^2}\). In this problem we use both \(u\)-substitution and integration by parts. First we write \(t^3 = t \cdot t^2\) and consider the indefinite integral \[\begin{equation*} \int t \cdot t^2 \cdot e^{t^2} \, dt\text{.} \end{equation*}\] We let \(z = t^2\) so that \(dz = 2t \, dt\), and thus \(t \, dt = \frac{1}{2} \, dz\). (We are using the variable \(z\) to perform a “\(z\)-substitution” first so that we may then apply integration by parts.) Under this \(z\)-substitution, we now have \[\begin{equation*} \int t \cdot t^2 \cdot e^{t^2} \, dt = \int z \cdot \frac{1}{2} e^{z} \, dz\text{.} \end{equation*}\] The resulting integral can be evaluated by parts.

These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables.

Activity

Evaluate each of the following indefinite integrals, using the provided hints.

  1. Evaluate \(\int \ln(x) \, dx\) by using integration by parts.
  2. Use the substitution \(z = t^2\) to transform the integral \(\int t^3 e^{t^2} \, dt\) to a new integral in the variable \(z\), and evaluate that new integral by parts.

3.3.3 Exercises

These are selected exercises from H. Jerome Keisler’s Elementary Calculus: An Infinitesimal Approach. Evaluate the integrals below. It may be necessary to use every integration rule and approach available. You may also have to resort to rewriting the expression in some way.

  1. Page 396, Item 15: \(\displaystyle\int \frac{x^3}{\sqrt{x^2-1}}\,dx\)
  2. Page 396, Item 27: \(\displaystyle\int t^3\sqrt{t^2+4}\,dt\)
  3. Page 396, Item 21: \(\displaystyle\int \frac{x^3}{\sqrt{x^2+1}}\,dx\)
  4. Page 396, Item 31: \(\displaystyle\int \frac{1}{x^3}\sqrt{\frac{1}{x}-1}\,dx\)
  5. Page 460, Item 31: \(\displaystyle\int \frac{2x}{x-1}\,dx\)
  6. Page 460, Item 35: \(\displaystyle\int \frac{dt}{t\ln t}\)
  7. Page 460, Item 39: \(\displaystyle\int x^n\ln x\,dx\), \(n\neq -1\)
  8. Page 460, Item 42: \(\displaystyle\int x\left(\ln x\right)^2\,dx\)
  9. Page 483, Example 1: \(\displaystyle\int \frac{dx}{\sqrt{x+1}-\sqrt{x}}\)
  10. Page 483, Example 3: \(\displaystyle\int\ln\left(\frac{x^2}{x+1}\right)\,dx\)
  11. Page 485, Item 7, \(\displaystyle\int \frac{dx}{x^{1/2}+1}\)
  12. Page 488, Item 10: \(\displaystyle\int \frac{e^{1/x}}{x^2}\,dx\)
  13. Page 488, Item 20: \(\displaystyle\int \frac{dx}{x^{1/3}+1}\)
  14. Page 488, Item 21: \(\displaystyle\int \ln\left(x^2+x^3\right)\,dx\)
  15. Page 488, Item 24: \(\displaystyle\int e^{\sqrt{x}}\,dx\)
  16. Page 488, Item 32: \(\displaystyle\int \frac{x\,dx}{\sqrt{x-1}-\sqrt{x}}\)
  17. Page 488, Item 49: \(\displaystyle\int \ln\left(x^2\sqrt{4x-1}\right)\,dx\)
  18. Page 489, Item 52: \(\displaystyle\int x^3 e^{x^2}\,dx\)
  19. Page 489, Item 54: \(\displaystyle\int \frac{x}{\sqrt{4x+1}}\,dx\)
  20. Page 489, Item 62: \(\displaystyle\int \ln\left(1+x^2\right)\,dx\)

3.4 When \(u\)-substitution and integration by parts fail to help

Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to find examples of functions for which neither technique produces an antiderivative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither \(u\)-substitution nor integration by parts proves fruitful for the indefinite integrals \[\begin{equation*} \int e^{x^2} \, dx \ \ \text{and} \ \ \int \sqrt{1+x^3} \, dx\text{.} \end{equation*}\] While there are other integration techniques, some of which we will consider briefly, none of them enables us to find an algebraic antiderivative for \(e^{x^2}\) or \(\sqrt{1+x^3}\). Antidifferentiation is much harder in general than differentiation.

We have learned two antidifferentiation techniques: \(u\)-substitution and integration by parts. The former is used to reverse the chain rule, while the latter to reverse the product rule. But we have seen that each works only in specialized circumstances. For example, while \(\int xe^{x^2} \, dx\) may be evaluated by \(u\)-substitution and \(\int x e^x \, dx\) by integration by parts, neither method provides a route to evaluate \(\int e^{x^2} \, dx\), and in fact an elementary algebraic antiderivative for \(e^{x^2}\) does not exist. No antidifferentiation method will provide us with a simple algebraic formula for a function \(F(x)\) that satisfies \(F'(x) = e^{x^2}\).

3.4.1 The method of partial fractions

The method of partial fractions is used to integrate rational functions. It involves reversing the process of finding a common denominator.

For example, let us evaluate \[\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx\text{.} \end{equation*}\]

If we factor the denominator, we can see how the integrand may be written as the sum of two fractions of the form \(\frac{A}{x-2} + \frac{B}{x+1}\), so we suppose that \[\begin{equation*} \frac{5x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \end{equation*}\] and look for the constants \(A\) and \(B\).

Multiplying both sides of this equation by \((x-2)(x+1)\), we find that \[\begin{equation*} 5x = A(x+1) + B(x-2)\text{.} \end{equation*}\]

Since we want this equation to hold for every value of \(x\), we can use insightful choices of specific \(x\)-values to help us find \(A\) and \(B\). Taking \(x = -1\), we have \[\begin{equation*} 5(-1) = A(0) + B(-3)\text{,} \end{equation*}\] so \(B = \frac{5}{3}\). Choosing \(x = 2\), it follows \[\begin{equation*} 5(2) = A(3) + B(0)\text{,} \end{equation*}\] so \(A = \frac{10}{3}\). Thus, \[\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx = \int \frac{10/3}{x-2} + \frac{5/3}{x+1} \, dx\text{.} \end{equation*}\]

This integral is straightforward to evaluate, and hence we find that \[\begin{equation*} \int \frac{5x}{x^2-x-2} \, dx = \frac{10}{3} \ln|x-2| + \frac{5}{3}\ln|x+1| + C\text{.} \end{equation*}\]

It turns out that we can use the method of partial fractions, together with \(u\)-substitution and other algebraic techniques, to rewrite any rational function \(R(x) = \frac{P(x)}{Q(x)}\) where the degree of the polynomial \(P\) is less than the degree of \(Q\) as a sum of simpler rational functions of one of the following forms: \[\begin{equation*} \frac{A}{x-c},\ \frac{A}{(x-c)^n},\ \frac{Ax+B}{x^2 + k},\ \text{or }\frac{Ax+B}{\left(x^2 + k\right)^n} \end{equation*}\] where \(A\), \(B\), and \(c\) are real numbers, and \(k\) is a positive real number. Because we can antidifferentiate each of these basic forms, partial fractions enables us to antidifferentiate any rational function.

Activity

For each of the following problems, evaluate the integral by the method of partial fractions.

  1. \(\displaystyle\int \frac{1}{x^2 - 2x - 3} \, dx\)
  2. \(\displaystyle\int \frac{x^2+1}{x^3 - x^2} \, dx\)
  3. Page 488 Item 33 of Keisler (which may need integration by parts, by substitution and by partial fractions): \(\displaystyle\int \frac{\ln x}{(1+x)^2}\,dx\)

3.4.2 Other options

Even the method of partial fractions you have seen just now may require even more wrangling. Consider the example where \[\begin{equation} \int \frac{x-2}{x^4 + x^2}\, dx, \end{equation}\] The partial fraction decomposition of the integrand is given by \[\begin{equation} \frac{x-2}{x^4 + x^2} = \frac{1}{x} - \frac{2}{x^2} + \frac{-x+2}{1+x^2}. \end{equation}\]

The last term looks easy to handle but requires much more knowledge about other basic antiderivatives than what you are exposed to. Observe that \[\begin{eqnarray}\int\frac{-x+2}{1+x^2}\, dx &=& -\int \frac{x}{1+x^2}\,dx+2\int\frac{1}{1+x^2}\,dx\\ &=& -\log(1+x^2)+2\int\frac{1}{1+x^2}\,dx.\end{eqnarray}\] Unfortunately, the usual approaches for \(u\)-substitution does not work. Integration by parts does not work either. But there are alternatives available:

  1. Use a table of integrals.
  2. Use some knowledge of derivatives of inverse trigonometric functions. In particular, \[\frac{d}{dx}\left[\tan^{-1}x\right]=\frac{1}{1+x^2}.\]
  3. Use another method of integration which uses a \(u\)-substitution involving trigonometric functions. For the case we have, we can use \(x=\tan\theta\).
  4. Use a computer algebra system or CAS.

A computer algebra system (CAS) is a computer program that is capable of executing symbolic mathematics, as opposed to numerical mathematics relying on approximations you have seen before in Chapter 1. For example, if we ask a CAS to solve the equation \(ax^2 + bx + c = 0\) for the variable \(x\), where \(a\), \(b\), and \(c\) are arbitrary constants, the program will return \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Research to develop the first CAS dates to the 1960s, and these programs became publicly available in the early 1990s. Two prominent examples are the programs Maple and Mathematica, which were among the first computer algebra systems to offer a graphical user interface. Today, Maple and Mathematica are exceptionally powerful professional software packages that can execute an amazing array of sophisticated mathematical computations. They are also very expensive, as each is a proprietary program. The CAS is an open-source, free alternative to Maple and Mathematica.

For the purposes of this text, when we need to use a CAS, we are going to turn instead to a similar, but somewhat different computational tool, the web-based “computational knowledge engine” called Wolfram Alpha. There are two features of Wolfram Alpha that make it stand out from the CAS options mentioned above: (1) unlike Maple and Mathematica, Wolfram Alpha is free (provided we are willing to navigate some pop-up advertising); and (2) unlike any of the three, the syntax in Wolfram Alpha is flexible. Think of Wolfram Alpha as being a little bit like doing a Google search: the program will interpret what is input, and then provide a summary of options.

While there is much to be enthusiastic about regarding CAS programs such as , there are several things we should be cautious about:

  1. A CAS only responds to exactly what is input.
  2. A CAS can answer using powerful functions from very advanced mathematics.
  3. There are problems that even a CAS cannot do without additional human insight.

The last two points are probably most important. Using sophisticated functions from more advanced mathematics is sometimes the way a CAS says to the user “I don’t know how to do this problem.” For example, if we want to evaluate \[\begin{equation*} \int e^{-x^2} \, dx\text{,} \end{equation*}\] and we ask to do so, the input integrate exp(-x^2) dx results in the output \[\begin{equation*} \int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\mathsf{erf} (x) + \text{constant}\text{.} \end{equation*}\]

The function \(\mathsf{erf}(x)\) is the error function, which is actually defined by an integral: \[\begin{equation*} \mathsf{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt\text{.} \end{equation*}\]

So, in producing output involving an integral, the CAS has basically reported back to us the very question we asked.

Finally, as remarked at (3) above, there are times that a CAS will actually fail without some additional human insight. If we consider the integral \[\begin{equation*} \int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx \end{equation*}\] and ask Wolfram Alpha to evaluate integrate (1+x) * exp(x) * sqrt(1+x^2) * exp(2x)) dx, the program thinks for a moment and then reports “no result found in terms of standard mathematical functions”. But in fact this integral is not that difficult to evaluate. If we let \(u = xe^{x}\), then \(du = (1+x)e^x \, dx\), which means that the preceding integral has form \[\begin{equation*} \int (1+x)e^x \sqrt{1+x^2e^{2x}} \, dx = \int \sqrt{1+u^2} \, du\text{,} \end{equation*}\] which is a straightforward one for any CAS to evaluate.

So, we should proceed with some caution: while any CAS is capable of evaluating a wide range of integrals (both definite and indefinite), there are times when the result can mislead us. We must think carefully about the meaning of the output, whether it is consistent with what we expect, and whether or not it makes sense to proceed.

3.5 Improper integrals

In this section, we ask the following questions:

  1. What are improper integrals and why are they important?
  2. What does it mean to say that an improper integral converges or diverges?
  3. What are some typical improper integrals that we can classify as convergent or divergent?

An important application of the definite integral measures the likelihood of certain events.2 For instance, consider a company that manufactures incandescent light bulbs. Based on a large volume of test results, they have determined3 that the fraction of light bulbs that fail between times \(t = a\) and \(t = b\) of use (where \(t\) is measured in months) is given by \[\begin{equation*} \int_a^b 0.3 e^{-0.3t} \, dt\text{.} \end{equation*}\]

For example, the fraction of light bulbs that fail during their third month of use is given by \[\begin{align*} \int_2^3 0.3e^{-0.3t} \, dt& = -e^{-0.3t} \bigg \vert_2^3\\ & = -e^{-0.9} + e^{-0.6}\\ & \approx 0.1422\text{.} \end{align*}\]

Thus about 14.22% of all lightbulbs fail between \(t = 2\) and \(t = 3\). Clearly we could adjust the limits of integration to measure the fraction of light bulbs that fail during any time period of interest.

Preview Activity

A company with a large customer base has a call center that receives thousands of calls a day. After studying the data that represents how long callers wait for assistance, they find that the function \(p(t) = 0.25e^{-0.25t}\) models the time customers wait in the following way: the fraction of customers who wait between \(t = a\) and \(t = b\) minutes is given by \[\begin{equation*} \int_a^b p(t) \, dt\text{.} \end{equation*}\]

Use this information to answer the following questions.

  1. Determine the fraction of callers who wait between 5 and 10 minutes.
  2. Determine the fraction of callers who wait between 10 and 20 minutes. c.Next, let’s study the fraction who wait up to a certain number of minutes:
    1. What is the fraction of callers who wait between 0 and 5 minutes?
    2. What is the fraction of callers who wait between 0 and 10 minutes?
    3. Between 0 and 15 minutes? Between 0 and 20?
  3. Let \(F(b)\) represent the fraction of callers who wait between \(0\) and \(b\) minutes. Find a formula for \(F(b)\) that involves a definite integral, and then use the FTC to find a formula for \(F(b)\) that does not involve a definite integral.
  4. What is the value of the limit \(\lim_{b \to \infty} F(b)\)? What is its meaning in the context of the problem?

3.5.1 Improper Integrals Involving Unbounded Intervals

In view of the above examples, we see that we may want to integrate over an interval whose upper limit grows without bound. For example, to find the fraction of light bulbs that fail eventually, we wish to find \[\begin{equation*} \lim_{b \to \infty} \int_0^b 0.3e^{-0.3t} \, dt\text{,} \end{equation*}\] for which we will also use the notation \[\begin{equation} \int_0^\infty 0.3e^{-0.3t} \, dt\text{.} \end{equation} \tag{3.6}\]

Such an integral can be interpreted as the area of an unbounded region, as pictured at right in Figure 3.4.

Figure 3.4: At left, the area bounded by \(p(t) = 0.3e^{-0.3t}\) on the finite interval \([0,b]\); at right, the result of letting \(b \to \infty\). By “\(\cdots\)” in the righthand figure, we mean that the region extends to the right without bound.

We call an integral for which the interval of integration is unbounded improper. For instance, the integrals \[\begin{equation*} \int_1^{\infty} \frac{1}{x^2} \, dx, \ \ \int_{-\infty}^0 \frac{1}{1+x^2} \, dx, \ \ \text{and} \int_{-\infty}^{\infty} e^{-x^2} \, dx \end{equation*}\] are all improper because they have limits of integration that involve \(\infty\). To evaluate an improper integral we replace it with a limit of proper integrals. That is, \[\begin{equation*} \int_0^\infty f(x) \, dx = \lim_{b \to \infty} \int_0^b f(x) \,dx\text{.} \end{equation*}\]

We first attempt to evaluate \(\int_0^b f(x) \,dx\) using the FTC, and then evaluate the limit. Is it even possible for the area of an unbounded region to be finite? The following activity explores this issue and others in more detail.

Activity

In this activity we explore the improper integrals \(\int_1^{\infty} \frac{1}{x} \, dx\) and \(\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\).

  1. First we investigate \(\int_1^{\infty} \frac{1}{x} \, dx\).
    1. Use the FTC to determine the exact values of \(\int_1^{10} \frac{1}{x} \, dx\), \(\int_1^{1000} \frac{1}{x} \, dx\), and \(\int_1^{100000} \frac{1}{x} \, dx\). Then, use your computational device to compute a decimal approximation of each result.
    2. Use the FTC to evaluate the definite integral \(\int_1^{b} \frac{1}{x} \, dx\) (which results in an expression that depends on \(b\)).
    3. Now, use your work from ii. to evaluate the limit given by \[\begin{equation*} \lim_{b \to \infty} \int_1^{b} \frac{1}{x} \, dx\text{.} \end{equation*}\]
  2. Next, we investigate \(\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\).
    1. Use the FTC to determine the exact values of \(\int_1^{10} \frac{1}{x^{3/2}} \, dx\), \(\int_1^{1000} \frac{1}{x^{3/2}} \, dx\), and \(\int_1^{100000} \frac{1}{x^{3/2}} \, dx\). Then, use your calculator to compute a decimal approximation of each result.
    2. Use the FTC to evaluate the definite integral \(\int_1^{b} \frac{1}{x^{3/2}} \, dx\) (which results in an expression that depends on \(b\)).
    3. Now, use your work from (ii.) to evaluate the limit given by \[\begin{equation*} \lim_{b \to \infty} \int_1^{b} \frac{1}{x^{3/2}} \, dx\text{.} \end{equation*}\]
  3. Plot the functions \(y = \frac{1}{x}\) and \(y = \frac{1}{x^{3/2}}\) on the same coordinate axes for the values \(x = 0 \ldots 10\). How would you compare their behavior as \(x\) increases without bound? What is similar? What is different?
  4. How would you characterize the value of \(\int_1^{\infty} \frac{1}{x} \, dx\)? of \(\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\)? What does this tell us about the respective areas bounded by these two curves for \(x \ge 1\)?

3.5.2 Convergence and Divergence

The previous activity suggests that \(\lim_{b \to \infty} \int_1^b f(x) \, dx\) is either finite or infinite (or it doesn’t exist). With these possibilities in mind, we introduce the following terminology.

Definition 3.1 (Convergence and divergence of improper integral) If \(f(x)\) is nonnegative for \(x \ge a\), then we say that the improper integral \(\int_a^{\infty} f(x) \, dx\) converges provided that \[\begin{equation*} \lim_{b \to \infty} \int_a^{b} f(x) \, dx \end{equation*}\] exists and is finite. Otherwise, we say that \(\int_a^{\infty} f(x) \, dx\) diverges.

We will restrict our interest to improper integrals for which the integrand is nonnegative. Also, we require that \(\lim_{x \to \infty} f(x) = 0\), for if \(f\) does not approach \(0\) as \(x \to \infty\), then it is impossible for \(\int_a^{\infty} f(x) \, dx\) to converge.

Activity

Determine whether each of the following improper integrals converges or diverges. For each integral that converges, find its exact value.

  1. \(\displaystyle \int_1^{\infty} \frac{1}{x^2} \, dx\)
  2. \(\displaystyle \int_0^{\infty} e^{-x/4} \, dx\)
  3. \(\displaystyle \int_2^{\infty} \frac{9}{(x+5)^{2/3}} \, dx\)
  4. \(\displaystyle \int_4^{\infty} \frac{3}{(x+2)^{5/4}} \, dx\)
  5. \(\displaystyle \int_0^{\infty} x e^{-x/4} \, dx\)
  6. \(\int_1^{\infty} \frac{1}{x^p} \, dx\), where \(p\) is a positive real number

3.5.3 Improper Integrals Involving Unbounded Integrands

An integral is also called improper if the integrand is unbounded on the interval of integration. For example, consider \[\begin{equation*} \int_0^1 \frac{1}{\sqrt{x}} \, dx\text{.} \end{equation*}\] Because \(f(x) = \frac{1}{\sqrt{x}}\) has a vertical asymptote at \(x = 0\), \(f\) is not continuous on \([0,1]\), and the integral represents the area of the unbounded region shown at right in Figure 3.5.

Figure 3.5: At left, the area bounded by \(f(x) = \frac{1}{\sqrt{x}}\) on the finite interval \([a,1]\); at right, the result of letting \(a \to 0^+\), where we see that the shaded region will extend vertically without bound.

We address the problem of the integrand being unbounded by replacing the improper integral with a limit of proper integrals. For example, to evaluate \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\), we replace \(0\) with \(a\) and let \(a\) approach 0 from the right. Thus, \[\begin{equation*} \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx\text{.} \end{equation*}\] We evaluate the proper integral \(\int_a^1 \frac{1}{\sqrt{x}} \, dx\), and then take the limit. We will say that the improper integral converges if this limit exists, and diverges otherwise. In this example, we observe that \[\begin{align*} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx\\ &= \lim_{a \to 0^+} \left. 2\sqrt{x}\, \right\vert_a^1\\ &= \lim_{a \to 0^+} 2\sqrt{1} - 2\sqrt{a}\\ &= 2\text{,} \end{align*}\] so the improper integral \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\) converges (to the value 2).

We have to be particularly careful with unbounded integrands, for they may arise in ways that may not initially be obvious. Consider, for instance, the integral \[\begin{equation*} \int_1^3 \frac{1}{(x-2)^2} \, dx\text{.} \end{equation*}\]

At first glance we might think that we can simply apply the FTC by antidifferentiating \(\frac{1}{(x-2)^2}\) to get \(-\frac{1}{x-2}\) and then evaluating from \(1\) to \(3\). Were we to do so, we would be erroneously applying the FTC because \(f(x) = \frac{1}{(x-2)^2}\) fails to be continuous throughout the interval, as seen in Figure 3.6.

Figure 3.6: The function \(f(x) = \frac{1}{(x-2)^2}\) on an interval including \(x = 2\).

Such an incorrect application of the FTC leads to an impossible result (\(-2\)), which would itself suggest that something we did must be wrong. Instead, we must address the vertical asymptote at \(x = 2\) by writing \[\begin{equation*} \int_1^3 \frac{1}{(x-2)^2} \, dx = \lim_{a \to 2^-} \int_1^a \frac{1}{(x-2)^2} \, dx + \lim_{b \to 2^+} \int_b^3 \frac{1}{(x-2)^2} \, dx\text{.} \end{equation*}\] We then evaluate two separate limits of proper integrals. For instance, doing so for the integral with \(a\) approaching \(2\) from the left, we find \[\begin{align*} \int_1^2 \frac{1}{(x-2)^2} \, dx&= \lim_{a \to 2^-} \int_1^a \frac{1}{(x-2)^2} \, dx\\ &= \lim_{a \to 2^-} -\frac{1}{(x-2)} \bigg\vert_1^a\\ &= \lim_{a \to 2^-} -\frac{1}{(a-2)} + \frac{1}{1-2}\\ &= \infty\text{,} \end{align*}\] since \(\frac{1}{a-2} \to -\infty\) as \(a\) approaches 2 from the left. Thus, the improper integral \(\int_1^2 \frac{1}{(x-2)^2} \, dx\) diverges; similar work shows that \(\int_2^3 \frac{1}{(x-2)^2} \, dx\) also diverges. From either of these two results, we can conclude that that the original integral, \(\int_1^3 \frac{1}{(x-2)^2} \, dx\) diverges, too.

Activity

For each of the following definite integrals, decide whether the integral is improper or not. If the integral is proper, evaluate it using the FTC. If the integral is improper, determine whether or not the integral converges or diverges; if the integral converges, find its exact value.

  1. \(\displaystyle \int_0^1 \frac{1}{x^{1/3}} \, dx\)
  2. \(\displaystyle \int_0^2 e^{-x} \, dx\)
  3. \(\displaystyle \int_1^4 \frac{1}{\sqrt{4-x}} \, dx\)
  4. \(\displaystyle \int_{-2}^2 \frac{1}{x^2} \, dx\)

3.5.4 Exercises

  1. Consider the following integrals and determine if the integrals are convergent or divergent. If these integrals are convergent, find their exact value.
    1. \(\displaystyle\int_{0}^{\,3} {\frac{9}{x\sqrt{x}}}\, dx\)
    2. \(\displaystyle \int_{2}^{\infty} 3x^{2}e^{-x^{3}} \,dx\)
    3. \(\displaystyle \int_{-\infty}^{-2}\,{e^{2 x}\over 1 + e^{2 x}}\,dx\)
    4. \({\displaystyle\int_{-2}^{2}{1\over v}\,dv}\)
    5. \(\displaystyle \int_e^{\infty} \frac{\ln(x)}{x} \, dx\)
    6. \(\displaystyle \int_e^{\infty} \frac{1}{x\ln(x)} \, dx\)
    7. \(\displaystyle \int_e^{\infty} \frac{1}{x(\ln(x))^2} \, dx\)
    8. \(\displaystyle \int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx\), where \(p\) is a positive real number
    9. \(\displaystyle \int_0^1 \frac{\ln(x)}{x} \, dx\)
    10. \(\displaystyle \int_0^1 \ln(x) \, dx\)
  2. Sometimes we may encounter an improper integral for which we cannot easily evaluate the limit of the corresponding proper integrals. For instance, consider \(\int_1^{\infty} \frac{1}{1+x^3} \, dx\). While it is hard (or perhaps impossible) to find an antiderivative for \(\frac{1}{1+x^3}\), we can still determine whether or not the improper integral converges or diverges by comparison to a simpler one. Observe that for all \(x \gt 0\), \(1 + x^3 \gt x^3\), and therefore \[\begin{equation*} \frac{1}{1+x^3} \lt \frac{1}{x^3}\text{.} \end{equation*}\] It therefore follows that \[\begin{equation*} \int_1^b \frac{1}{1+x^3} \, dx \lt \int_1^b \frac{1}{x^3} \, dx \end{equation*}\] for every \(b \gt 1\). If we let \(b \to \infty\) so as to consider the two improper integrals \(\int_1^\infty \frac{1}{1+x^3} \, dx\) and \(\int_1^\infty \frac{1}{x^3} \, dx\), we know that the larger of the two improper integrals converges. And thus, since the smaller one lies below a convergent integral, it follows that the smaller one must converge, too. In particular, \(\int_1^\infty \frac{1}{1+x^3} \, dx\) must converge, even though we never explicitly evaluated the corresponding limit of proper integrals. We use this idea and similar ones in the exercises that follow.
    1. Explain why \(x^2 + x + 1 \gt x^2\) for all \(x \ge 1\), and hence show that \(\int_1^{\infty} \frac{1}{x^2 + x + 1} \, dx\) converges by comparison to \(\int_1^{\infty} \frac{1}{x^2} \, dx\).
    2. Observe that for each \(x \gt 1\), \(\ln(x) \lt x\). Explain why \[\begin{equation*} \int_2^b \frac{1}{x} \, dx \lt \int_2^b \frac{1}{\ln(x)} \,dx \end{equation*}\] for each \(b \gt 2\). Why must it be true that \(\int_2^\infty \frac{1}{\ln(x)} \, dx\) diverges?
    3. Explain why \(\sqrt{\frac{x^4+1}{x^4}} \gt 1\) for all \(x \gt 1\). Then, determine whether or not the improper integral \[\begin{equation*} \int_1^{\infty} \frac{1}{x} \cdot \sqrt{\frac{x^4+1}{x^4}} \, dx \end{equation*}\] converges or diverges.

3.6 Summary of the main points

  1. We can find the exact value of a definite integral without taking the limit of a Riemann sum or using a familiar area formula by finding the antiderivative of the integrand, and hence applying the Fundamental Theorem of Calculus.

  2. The Fundamental Theorem of Calculus says that if \(f\) is a continuous function on \([a,b]\) and \(F\) is an antiderivative of \(f\), then \[\begin{equation*} \int_a^b f(x) \, dx = F(b) - F(a)\text{.} \end{equation*}\] Hence, if we can find an antiderivative for the integrand \(f\), evaluating the definite integral comes from simply computing the change in \(F\) on \([a,b]\).

  3. To find algebraic formulas for antiderivatives of more complicated algebraic functions, we need to think carefully about how we can reverse known differentiation rules. To that end, it is essential that we understand and recall known derivatives of basic functions, as well as the standard derivative rules.

  4. The indefinite integral provides notation for antiderivatives. When we write \(\int f(x) \, dx\), we mean the general antiderivative of \(f\). In particular, if we have functions \(f\) and \(F\) such that \(F' = f\), the following two statements say the exact thing: \[\begin{equation*} \frac{d}{dx}[F(x)] = f(x) \ \text{and} \ \int f(x) \, dx = F(x) + C\text{.} \end{equation*}\] That is, \(f\) is the derivative of \(F\), and \(F\) is an antiderivative of \(f\).

  5. The technique of \(u\)-substitution helps us to evaluate indefinite integrals of the form \(\int f(g(x))g'(x) \, dx\) through the substitutions \(u = g(x)\) and \(du = g'(x) \, dx\), so that \[\begin{equation*} \int f(g(x))g'(x) \, dx = \int f(u) \, du\text{.} \end{equation*}\] A key part of choosing the expression in \(x\) to be represented by \(u\) is the identification of a function-derivative pair. To do so, we often look for an ``inner’’ function \(g(x)\) that is part of a composite function, while investigating whether \(g'(x)\) (or a constant multiple of \(g'(x)\)) is present as a multiplying factor of the integrand.

  6. Through the method of integration by parts, we can evaluate indefinite integrals that involve products of basic functions. Using a substitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an effort to find a different product of functions that is easier to integrate.

  7. If the algebraic structure of an integrand is a product of basic functions in the form \(\int f(x) g'(x) \, dx\), we can use the substitution \(u = f(x)\) and \(dv = g'(x) \,dx\) and apply the rule \[\begin{equation*} \int u \, dv = uv - \int v \, du \end{equation*}\] to evaluate the original integral \(\int f(x) g'(x) \, dx\) by instead evaluating% \[\begin{equation*} \int v \, du = \int f'(x) g(x) \, dx\text{.} \end{equation*}\]

  8. When deciding to integrate by parts, we have to select both \(u\) and \(dv\). That selection is guided by the overall principle that the new integral \(\int v \, du\) not be more difficult than the original integral \(\int u \, dv\). In addition, it is often helpful to recognize if one of the functions present is much easier to differentiate than antidifferentiate (such as \(\ln(x)\)), in which case that function often is best assigned the variable \(u\). In addition, \(dv\) must be a function that we can antidifferentiate.

  9. We can antidifferentiate any rational function with the method of partial fractions. Any polynomial function can be factored into a product of linear and irreducible quadratic terms, so any rational function may be written as the sum of a polynomial plus rational terms of the form \(\frac{A}{(x-c)^n}\) (where \(n\) is a natural number) and \(\frac{Bx+C}{x^2 + k}\) (where \(k\) is a positive real number).

  10. Computer algebra systems can play an important role in finding antiderivatives, though we must be cautious to use correct input, to watch for unusual or unfamiliar advanced functions that the CAS may cite in its result, and to consider the possibility that a CAS may need further assistance or insight from us in order to answer a particular question.

  11. An integral \(\int_a^b f(x) \, dx\) can be improper if at least one of \(a\) or \(b\) is \(\pm \infty\), making the interval unbounded, or if \(f\) has a vertical asymptote at \(x = c\) for some value of \(c\) that satisfies \(a \le c \le b\). One reason that improper integrals are important is that certain probabilities can be represented by integrals that involve infinite limits.

  12. When we encounter an improper integral, we work to understand it by replacing the improper integral with a limit of proper integrals. For instance, we write \[\begin{equation*} \int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx\text{,} \end{equation*}\] and then work to determine whether the limit exists and is finite. For any improper integral, if the resulting limit of proper integrals exists and is finite, we say the improper integral converges. Otherwise, the improper integral diverges.

  13. An important class of improper integrals is given by \[\begin{equation*} \int_1^{\infty} \frac{1}{x^p} \, dx \end{equation*}\] where \(p\) is a positive real number. We can show that this improper integral converges whenever \(p \gt 1\), and diverges whenever \(0 \lt p \le 1\). A related class of improper integrals is \(\int_0^1 \frac{1}{x^p} \, dx\), which converges for \(0 \lt p \lt 1\), and diverges for \(p \ge 1\).

  1. Recall that the general antiderivative of a function includes “\(+C\)” to reflect the entire family of functions that share the same derivative.↩︎

  2. We will discuss some more details about this application area in Chapter 4.↩︎

  3. You have to wonder how the company arrived at the resulting expression.↩︎