Download Integration Techniques: Substitution and Integration by Parts and more Study notes Calculus in PDF only on Docsity! LECTURE THIRTEEN Recall the following result from previous lectures. Corollary 1. Let f(t) be a continuous function [a, b]. If G is an antiderivative of f on [a, b], then b ∫ a f(t)dt = [G(x)]ba = G(b) − G(a). Example: Let m, n be positive integers such that m n 6= −1 and n 6= 0. Then, Since G(x) = ( 1 m n +1 ) x m n +1 is an antiderivative of the function f(x) = x m n , we have that 2 ∫ 1 x m n dx = [( 1 m n + 1 ) x m n +1 ]2 1 = ( 1 m n + 1 ) 2 m n +1− ( 1 m n + 1 ) 1 m n +1 = ( 1 m n + 1 ) (2 m n +1−1). So far we only know the following antiderivatives. (1) ∫ αdx = αx + c, where α is a constant. (2) ∫ x m n dx = 1m n +1 x m n +1 + c. (3) ∫ 1 x dx = ln |x| + c. (4) ∫ exdx = ex + c. Using the above known antiderivatives, we can find the antiderivatives of other func- tions, for example, ∫ (2x + 5 + ex)dx = x2 + 5x + ex + C. So, we have that 1 ∫ 0 2x + 5 + exdx = [ x2 + 5x + ex ]1 0 = (12 + 5(1) + e1) − (02 + 5(0) + e0) = 5 + e. 1. Substitution: Change of Variables Let g be a continuous function on [a, b] such that the derivative g′ is also continuous on [a, b]. Let f be a function which is continuous on g([a, b]). Then, (1.1) b ∫ x=a f(g(x))g′(x)dx = g(b) ∫ u=g(a) f(u)du, 1 2 LECTURE THIRTEEN Example: Evaluate 2 ∫ 1 2x √ x2 − 1dx. We apply the formula (1.1). Let g(x) = x2 − 1, and let f(x) = √x. Then we have that 2 ∫ 1 2x √ x2 − 1dx = 2 ∫ 1 f(g(x))g′(x)dx. Also, F (x) = 2 3 x 3 2 is an antiderivative for f(x) = √ x. Note that, g(1) = 0, g(2) = 3, and g′(x) = 2x So by applying the formula (1.1), we get 2 ∫ 1 2x √ x2 − 1dx = 2 ∫ x=1 f(g(x))g′(x)dx = 3 ∫ u=0 f(u)du = 3 ∫ u=0 u 1 2 du = [ 2 3 u 3 2 ]3 0 = 2 3 [ 3 3 2 − 0 32 ] = 2 3 3 3 2 . In practice, If we have to find an integral of the form (1.2) b ∫ x=a f(g(x))g′(x)dx, then it is helpful to follow the following steps. We let u = g(x), and then du = g′(x)dx, and we change the limits of integration a and b by g(a) and g(b). Then the integral equals g(b) ∫ u=g(a) f(u)du. Example: Evaluate 2 ∫ x=1 2xex 2 dx Let u = x2. Then du = 2xdx, and 2 ∫ x=1 2xex2dx = 4 ∫ u=1 eudu = [eu]41 = e 4 − e1 = e4 − e. Remark: The above method can also be used to find antiderivatives. Note that an antiderivative of a function of the form f(g(x))g′(x) is F (g(x) where F (x) is an antiderivative of f(x). This follows from the chain rule. We write this in the following way. ∫ f(g(x))g′(x)dx = F (g(x) + c. In practice it is helpful to follow the steps below in order to find the antiderivative ∫ f(g(x))g′(x)dx. We let u = g(x) and then du = g′(x)dx, to get ∫ f(g(x))g′(x)dx = ∫ f(u)du = F (u) + c = F (g(x)) + c.
