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Substitution for Definite Integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

Theorem 5.8

Substitution with Definite Integrals

Let \(u = g \left(\right. x \left.\right)\) and let \(g^{'}\) be continuous over an interval \(\left[\right. a , b \left]\right. ,\) and let f be continuous over the range of \(u = g \left(\right. x \left.\right) .\) Then,

\[\int_{a}^{b} f \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} ( x \left.\right) d x = \int_{g \left(\right. a \left.\right)}^{g \left(\right. b \left.\right)} f \left(\right. u \left.\right) d u .\]

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if \(F \left(\right. x \left.\right)\) is an antiderivative of \(f \left(\right. x \left.\right) ,\) we have

\[\int f \left(\right. g \left(\right. x \left.\right) \left.\right) g^{'} ( x \left.\right) d x = F \left(\right. g \left(\right. x \left.\right) \left.\right) + C .\]

Then

\[\begin{aligned} \int_{a}^{b} f \left[\right. g \left(\right. x \left.\right) \left]\right. g^{'} ( x \left.\right) d x & = \left(F \left(\right. g \left(\right. x \left.\right) \left.\right) \left|\right.\right)_{x = a}^{x = b} \\ & = F \left(\right. g \left(\right. b \left.\right) \left.\right) - F \left(\right. g \left(\right. a \left.\right) \left.\right) \\ & = \left(F \left(\right. u \left.\right) \left|\right.\right)_{u = g \left(\right. a \left.\right)}^{u = g \left(\right. b \left.\right)} \\ \\ \\ & = \int_{g \left(\right. a \left.\right)}^{g \left(\right. b \left.\right)} f \left(\right. u \left.\right) d u , \end{aligned}\]

and we have the desired result.

Example 5.34

Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate \(\int_{0}^{1} x^{2} \left(\right. 1 + 2 x^{3} \left.\right)^{5} d x .\)

Solution

Let \(u = 1 + 2 x^{3} ,\) so \(d u = 6 x^{2} d x .\) Since the original function includes one factor of x2 and \(d u = 6 x^{2} d x ,\) multiply both sides of the du equation by \(1 / 6 .\) Then,

\[\begin{aligned} d u & = & 6 x^{2} d x \\ \frac{1}{6} d u & = & x^{2} d x . \end{aligned}\]

To adjust the limits of integration, note that when \(x = 0 , u = 1 + 2 \left(\right. 0 \left.\right) = 1 ,\) and when \(x = 1 , u = 1 + 2 \left(\right. 1 \left.\right) = 3 .\) Then

\[\int_{0}^{1} x^{2} \left(\right. 1 + 2 x^{3} \left.\right)^{5} d x = \frac{1}{6} \int_{1}^{3} u^{5} d u .\]

Evaluating this expression, we get

\[\begin{aligned} \\ \\ \frac{1}{6} \int_{1}^{3} u^{5} d u & = \left(\right. \frac{1}{6} \left.\right) \left(\right. \frac{u^{6}}{6} \left.\right) \left|\right._{1}^{3} \\ & = \frac{1}{36} \left[\right. \left(\right. 3 \left.\right)^{6} - \left(\right. 1 \left.\right)^{6} \left]\right. \\ & = \frac{182}{9} . \end{aligned}\]

Checkpoint 5.29

Use substitution to evaluate the definite integral \(\int_{−1}^{0} y \left(\right. 2 y^{2} - 3 \left.\right)^{5} d y .\)

Example 5.35

Using Substitution with an Exponential Function

Use substitution to evaluate \(\int_{0}^{1} x e^{4 x^{2} + 3} d x .\)

Solution

Let \(u = 4 x^{2} + 3 .\) Then, \(d u = 8 x d x .\) To adjust the limits of integration, we note that when \(x = 0 , u = 3 ,\) and when \(x = 1 , u = 7 .\) So our substitution gives

\[\begin{aligned} \int_{0}^{1} x e^{4 x^{2} + 3} d x & = \frac{1}{8} \int_{3}^{7} e^{u} d u \\ \\ & = \frac{1}{8} e^{u} \left|\right._{3}^{7} \\ & = \frac{e^{7} - e^{3}}{8} \\ & \approx 134.568. \end{aligned}\]

Checkpoint 5.30

Use substitution to evaluate \(\int_{0}^{1} x^{2} \text{cos} \left(\right. \frac{\pi}{2} x^{3} \left.\right) d x .\)

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example 5.36.

Example 5.36

Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate \(\int_{0}^{\pi / 2} \text{cos}^{2} \theta d \theta .\)

Solution

Let us first use a trigonometric identity to rewrite the integral. The trig identity \(\text{cos}^{2} \theta = \frac{1 + \text{cos} 2 \theta}{2}\) allows us to rewrite the integral as

\[\int_{0}^{\pi / 2} \text{cos}^{2} \theta d \theta = \int_{0}^{\pi / 2} \frac{1 + \text{cos} 2 \theta}{2} d \theta .\]

Then,

\[\begin{aligned} \int_{0}^{\pi / 2} \left(\right. \frac{1 + \text{cos} 2 \theta}{2} \left.\right) d \theta & = \int_{0}^{\pi / 2} \left(\right. \frac{1}{2} + \frac{1}{2} \text{cos} 2 \theta \left.\right) d \theta \\ \\ \\ & = \frac{1}{2} \int_{0}^{\pi / 2} d \theta + \frac{1}{2} \int_{0}^{\pi / 2} \text{cos} 2 \theta d \theta . \end{aligned}\]

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let \(u = 2 \theta .\) Then, \(d u = 2 d \theta ,\) or \(\frac{1}{2} d u = d \theta .\) Also, when \(\theta = 0 , u = 0 ,\) and when \(\theta = \pi / 2 , u = \pi .\) Expressing the second integral in terms of u, we have

\[\begin{aligned} \\ \\ \frac{1}{2} \int_{0}^{\pi / 2} d \theta + \frac{1}{2} \int_{0}^{\pi / 2} \text{cos} 2 \theta d \theta & = \frac{1}{2} \int_{0}^{\pi / 2} d \theta + \frac{1}{2} \left(\right. \frac{1}{2} \left.\right) \int_{0}^{\pi} \text{cos} u d u \\ & = \frac{\theta}{2} \left|\right._{\theta = 0}^{\theta = \pi / 2} + \frac{1}{4} \text{sin} u \left|\right._{u = 0}^{u = \pi} \\ & = \left(\right. \frac{\pi}{4} - 0 \left.\right) + \left(\right. 0 - 0 \left.\right) = \frac{\pi}{4} . \end{aligned}\]

This lesson is part of:

Integration

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