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INTEGRALS 5

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Indefinite Integrals INTEGRALS

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The notation ∫ f(x) dx is traditionally used for an antiderivative of f and is called an indefinite integral. Thus, ∫ f(x) dx = F(x) means F’(x) = f(x) INDEFINITE INTEGRAL

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INDEFINITE INTEGRALS For example, we can write Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).

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INDEFINITE INTEGRALS Any formula can be verified by differentiating the function on the right side and obtaining the integrand. For instance,

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TABLE OF INDEFINITE INTEGRALS Table 1

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INDEFINITE INTEGRALS Thus, we write with the understanding that it is valid on the interval (0, ∞ ) or on the interval (- ∞, 0).

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INDEFINITE INTEGRALS This is true despite the fact that the general antiderivative of the function f(x) = 1/x 2, x ≠ 0, is:

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INDEFINITE INTEGRALS Find the general indefinite integral ∫ (10x 4 – 2 sec 2 x) dx Using our convention and Table 1, we have: ∫(10x 4 – 2 sec 2 x) dx = 10 ∫ x 4 dx – 2 ∫ sec 2 x dx = 10(x 5 /5) – 2 tan x + C = 2x 5 – 2 tan x + C You should check this answer by differentiating it. Example 1

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INDEFINITE INTEGRALS Evaluate This indefinite integral isn’t immediately apparent in Table 1. So, we use trigonometric identities to rewrite the function before integrating: Example 2

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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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Our 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 = 1 + x 2 be the quantity under the root sign in the integral below, Then, the differential of u is du = 2x dx. INTRODUCTION

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Notice that, if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in 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 INTRODUCTION

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In general, this method works whenever we have an integral that we can write in the form ∫ f(g(x))g’(x) dx INTRODUCTION

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Observe that, if F’ = f, then ∫ F’(g(x))g’(x) dx = F(g(x)) + C because, by the Chain Rule, Equation 3 INTRODUCTION

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

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Writing F’ = f, we get: ∫ f(g(x))g’(x) dx = ∫ f(u) du Thus, we have proved the following rule. 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 Thus, the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were 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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SUBSTITUTION RULE 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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SUBSTITUTION RULE The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable x to a new variable u that is a function of x. Thus, in Example 1, we replaced the integral ∫ x 3 cos(x 4 + 2) dx by the simpler integral ¼ ∫ cos u du.

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SUBSTITUTION RULE The main challenge in using the rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor. This was the case in Example 1.

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SUBSTITUTION RULE If that is not possible, try choosing u to be some complicated part of the integrand— perhaps the inner function in a composite function.

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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: SUBSTITUTION RULE E. g. 2—Solution 1

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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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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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