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INTEGRALS 5

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5.5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function, making integration easier. INTEGRALS

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Antidifferentiation formulas don’t tell us how to evaluate integrals such as Equation 1 INTRODUCTION

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To find this integral, we use the problem- solving strategy of introducing something extra. The ‘something extra’ is a new variable. We change from the variable x to a new variable u. INTRODUCTION

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Suppose we let u be the quantity under the root sign in Equation 1, u = 1 + x 2. Then, the differential of u is du = 2x dx. INTRODUCTION

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So, formally, without justifying our calculation, we could write: Equation 2 INTRODUCTION

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However, now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2: INTRODUCTION

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Observe that, if F’ = f, then because, by the Chain Rule, Equation 3 INTRODUCTION

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If we make the ‘change of variable’ or ‘substitution’ u = g(x), from Equation 3, we have: INTRODUCTION

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SUBSTITUTION RULE If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ f(g(x))g’(x) dx = ∫ f(u) du Equation 4

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SUBSTITUTION RULE Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that, if u = g(x), then du = g’(x) dx. So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.

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SUBSTITUTION RULE Find ∫ x 3 cos(x 4 + 2) dx We make the substitution u = x 4 + 2. This is because its differential is du = 4x 3 dx, which, apart from the constant factor 4, occurs in the integral. Example 1

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Solution: Thus, using x 3 dx = du/4 and the Substitution Rule, we have: Notice that, at the final stage, we had to return to the original variable x. Example 1

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Finding the right substitution is a bit of an art. It’s not unusual to guess wrong. If your first guess doesn’t work, try another substitution. SUBSTITUTION RULE

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Evaluate Let u = 2x + 1. Then, du = 2 dx. So, dx = du/2. E. g. 2—Solution 1

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Thus, the rule gives: Solution1: E. g. 2—Solution 1

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Another possible substitution is Then, So, Alternatively, observe that u 2 = 2x + 1. So, 2u du = 2 dx. Solution2: E. g. 2—Solution 2

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Thus, Solution2: E. g. 2—Solution 2

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SUBSTITUTION RULE Find Let u = 1 – 4x 2. Then, du = -8x dx. So, x dx = -1/8 du and Example 3

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Solution: Here, we have used a computer to graph both the integrand and its indefinite integral We take the case C = 0. Figure 5.5.1, p. 335

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Solution: Notice that g(x): Decreases when f(x) is negative Increases when f(x) is positive Has its minimum value when f(x) = 0 Figure 5.5.1, p. 335

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SUBSTITUTION RULE Calculate ∫ cos 5x dx If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore, Example 4

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SUBSTITUTION RULE Find An appropriate substitution becomes more obvious if we factor x 5 as x 4. x. Let u = 1 + x 2. Then, du = 2x dx. So, x dx = du/2. Example 5

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Solution: Also, x 2 = u – 1; so, x 4 = (u – 1) 2 : Example 5

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Evaluate the indefinite integral first and then use the FTC for definite integral. For instance, using the result of Example 2, we have: DEFINITE INTEGRALS

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If g’ is continuous on [a, b] and f is continuous on the range of u = g(x), then SUB. RULE FOR DEF. INTEGRALS Equation 5

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Let F be an antiderivative of f. Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x). So, by Part 2 of the FTC (FTC2), we have: SUB. RULE FOR DEF. INTEGRALS Proof

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However, applying the FTC2 a second time, we also have: SUB. RULE FOR DEF. INTEGRALS Proof

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Evaluate using Equation 5. Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2 Example 6 SUB. RULE FOR DEF. INTEGRALS Figure 5.5.2a, p. 337

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To find the new limits of integration, we note that: When x = 0, u = 2(0) + 1 = 1 When x = 4, u = 2(4) + 1 = 9 Example 6 Solution: Figure 5.5.2b, p. 337

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Thus, Example 6 Solution:

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Evaluate Let u = 3 – 5x. Then, du = –5 dx, so dx = –du/5. When x = 1, u = –2, and when x = 2, u = –7. Example 7 SUB. RULE FOR DEF. INTEGRALS

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Thus, Example 7 Solution:

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SYMMETRY The next theorem uses the Substitution Rule for Definite Integrals to simplify the calculation of integrals of functions that possess symmetry properties.

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INTEGS. OF SYMM. FUNCTIONS Suppose f is continuous on [–a, a]. a.If f is even, [f(–x) = f(x)], then b.If f is odd, [f(-x) = -f(x)], then Theorem 6

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We split the integral in two: Proof—Equation 7 INTEGS. OF SYMM. FUNCTIONS

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In the first integral in the second part, we make the substitution u = –x. Then, du = – dx, and when x = – a, u = a. INTEGS. OF SYMM. FUNCTIONS Proof

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Therefore, Proof INTEGS. OF SYMM. FUNCTIONS

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So, Equation 7 becomes: Proof—Equation 8 INTEGS. OF SYMM. FUNCTIONS

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If f is even, then f(–u) = f(u). So, Equation 8 gives: INTEGS. OF SYMM. FUNCTIONS Proof a

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If f is odd, then f(–u) = –f(u). So, Equation 8 gives: INTEGS. OF SYMM. FUNCTIONS Proof b

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Theorem 6 is illustrated here. INTEGS. OF SYMM. FUNCTIONS Figure 5.5.3, p. 338

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For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry. INTEGS. OF SYMM. FUNCTIONS Figure 5.5.3a, p. 338

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Therefore, part (b) says the integral is 0 because the areas cancel. INTEGS. OF SYMM. FUNCTIONS Figure 5.5.3b, p. 338

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As f(x) = x 6 + 1 satisfies f(–x) = f(x), it is even. So, Example 8 INTEGS. OF SYMM. FUNCTIONS

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As f(x) = (tan x)/ (1 + x 2 + x 4 ) satisfies f( – x) = – f(x), it is odd. So, Example 9 INTEGS. OF SYMM. FUNCTIONS

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