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5 INTEGRALS INTEGRALS Indefinite Integrals INDEFINITE INTEGRAL 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 INTEGRALS For example, we can write 3 x d x 2 2 x dx 3 C because dx 3 C x 3 Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C). INDEFINITE INTEGRALS Any formula can be verified by differentiating the function on the right side and obtaining the integrand. For instance, sec 2 x dx tan x C because d (tan x C ) sec 2 x dx TABLE OF INDEFINITE INTEGRALS cf ( x) dx c f ( x) dx Table 1 [ f ( x) g ( x)] dx f ( x) dx g ( x) dx k dx kx C n 1 x n x dx n 1 C (n 1) sin x dx cos x C cos x dx sin x C sec csc 2 x dx tan x C sec x tan x dx sec x C 2 x dx cot x C csc x cot x dx csc x C INDEFINITE INTEGRALS Thus, we write 1 1 x 2 dx x C with the understanding that it is valid on the interval (0, ∞) or on the interval (-∞, 0). INDEFINITE INTEGRALS This is true despite the fact that the general antiderivative of the function f(x) = 1/x2, x ≠ 0, is: 1 C if x 0 1 x F ( x) 1 C if x 0 2 x INDEFINITE INTEGRALS Example 1 Find the general indefinite integral ∫ (10x4 – 2 sec2x) dx Using our convention and Table 1, we have: ∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx = 10(x5/5) – 2 tan x + C = 2x5 – 2 tan x + C You should check this answer by differentiating it. INDEFINITE INTEGRALS Example 2 cos Evaluate sin 2 d This indefinite integral isn’t immediately apparent in Table 1. So, we use trigonometric identities to rewrite the function before integrating: cos 1 cos sin 2 d sin sin d csc cot d csc C INTEGRALS 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. Equation 1 INTRODUCTION • Our antidifferentiation formulas don’t tell us how to evaluate integrals such as 2 x 1 x dx 2 INTRODUCTION • To find this integral, we use the problemsolving strategy of introducing something extra. The ‘something extra’ is a new variable. We change from the variable x to a new variable u. INTRODUCTION • Suppose we let u = 1 + x2 be the quantity under the root sign in the integral below, 2x 1 x dx 2 Then, the differential of u is du = 2x dx. INTRODUCTION • 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 2 x 1 x dx 2 Equation 2 INTRODUCTION • So, formally, without justifying our calculation, we could write: 2 x 1 x dx 1 x 2 x dx 2 2 udu u 2 3 3/ 2 C ( x 1) 2 3 2 3/ 2 C INTRODUCTION • However, now we can check that we have the correct answer by using the Chain Rule to differentiate d 2 2 3 ( x 1)3 2 C 32 32 ( x 2 1)1 2 2 x dx 2x x 1 2 INTRODUCTION • In general, this method works whenever we have an integral that we can write in the form ∫ f(g(x))g’(x) dx INTRODUCTION Equation 3 • Observe that, if F’ = f, then • ∫ F’(g(x))g’(x) dx = F(g(x)) + C • because, by the Chain Rule, d F ( g ( x)) F '( g ( x)) g '( x) dx INTRODUCTION • If we make the ‘change of variable’ or ‘substitution’ u = g(x), we have: F '( g ( x )) g '( x ) dx F ( g ( x )) C F (u ) C F '(u ) du INTRODUCTION • Writing F’ = f, we get: ∫ f(g(x))g’(x) dx = ∫ f(u) du Thus, we have proved the following rule. SUBSTITUTION RULE Equation 4 • 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 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. SUBSTITUTION RULE • Thus, the Substitution Rule says: • It is permissible to operate with dx and du after integral signs as if they were differentials. SUBSTITUTION RULE Example 1 • Find ∫ x3 cos(x4 + 2) dx We make the substitution u = x4 + 2. This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral. Example 1 SUBSTITUTION RULE • Thus, using x3 dx = du/4 and the Substitution Rule, we have: 1 1 x cos( x 2) dx cos u du 4 4 cos udu 3 4 14 sin u C sin( x 2) C 1 4 4 Notice that, at the final stage, we had to return to the original variable x. 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 ∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du. 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. 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. SUBSTITUTION RULE • 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. E. g. 2—Solution 1 SUBSTITUTION RULE • Evaluate Let u = 2x + 1. Then, du = 2 dx. So, dx = du/2. 2 x 1 dx E. g. 2—Solution 1 SUBSTITUTION RULE Thus, the rule gives: 2 x 1 dx 1 2 du u 2 u 12 du 32 1 u C 2 3/ 2 1 32 3u C (2 x 1) 1 3 32 C Example 3 SUBSTITUTION RULE • Find x 1 4x2 dx Let u = 1 – 4x2. Then, du = -8x dx. So, x dx = -1/8 du and x 1 4x 2 dx 1 8 1 du 18 u 1 2 du u 18 (2 u ) C 14 1 4 x 2 C SUBSTITUTION RULE Example 4 • Calculate ∫ cos 5x dx If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore, 1 cos 5 x dx 5 cos u du 15 sin u C 15 sin 5 x C