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8-2: Integration By Parts Objectives: 1. To find an antiderivative using integration by parts Assignment: • P. 531-533: 9-17 odd, 23-33 odd, 39, 43, 49, 53, 73-78, 99, 103a-c, 107 • Homework Supplement Warm Up Let 𝑢 and 𝑣 be differentiable functions of 𝑥. 𝑑 Find 𝑢𝑣 . 𝑑𝑥 𝑑 𝑢𝑣 = 𝑣𝑢′ + 𝑢𝑣′ 𝑑𝑥 𝑑 𝑑𝑢 𝑑𝑣 𝑢𝑣 = 𝑣 +𝑢 𝑑𝑥 𝑑𝑥 𝑑𝑥 Product Rule The product of two differentiable functions 𝑓 and 𝑔 is also differentiable such that 𝑑 𝑓(𝑥) ∙ 𝑔(𝑥) = 𝑓 ′ 𝑥 ∙ 𝑔 𝑥 + 𝑓(𝑥) ∙ 𝑔′ (𝑥) 𝑑𝑥 Derivative of the first times the second + first times the derivative of the second Objective 1 You will be able to find an antiderivative using integration by parts Working Backwards Given that 𝑢 and 𝑣 are differentiable functions of 𝑥, let’s work the Product Rule backwards. Solve for 𝑢 𝑑𝑣 𝑑 𝑑𝑢 𝑑𝑣 𝑢𝑣 = 𝑣 +𝑢 𝑑𝑥 𝑑𝑥 𝑑𝑥 Multiply by 𝑑𝑥 Integrate both sides 𝑑 𝑢𝑣 = 𝑣 𝑑𝑢 + 𝑢 𝑑𝑣 𝑑 𝑢𝑣 = 𝑢𝑣 = 𝑣 𝑑𝑢 + 𝑢 𝑑𝑣 𝑣 𝑑𝑢 + 𝑢 𝑑𝑣 = 𝑢𝑣 − 𝑢 𝑑𝑣 𝑣 𝑑𝑢 Integration By Parts If 𝑢 and 𝑣 are functions of 𝑥 and have continuous derivatives, then 𝑢 𝑑𝑣 = 𝑢𝑣 − Try letting 𝑢 be the part whose derivative is simpler than 𝑢. 𝑣 𝑑𝑢 Try letting 𝑑𝑣 be the more complicated part. Notice that you are exchanging 𝑢 𝑑𝑣 for 𝑣 𝑑𝑢, which should be simpler and easier to integrate. Mnemonic Device If 𝑢 and 𝑣 are functions of 𝑥 and have continuous derivatives, then 𝑢 𝑑𝑣 Try letting 𝑢 be the part whose derivative is simpler than 𝑢. 𝑢←L I on g v = 𝑢𝑣 − 𝑣 𝑑𝑢 ae r r i s t e Try letting 𝑑𝑣 be h the more mt complicated part. i r c i g P o w e r E x p o n e n t i a l T → 𝑑𝑣 r i g o n o m e t r i c We use integration by parts when the integrand is a product, usually of an algebraic times a transcendental function. Organizational Box If 𝑢 and 𝑣 are functions of 𝑥 and have continuous derivatives, then Original New Differentiate 𝑢= 𝑣 𝑑𝑢 𝑣= 𝑑𝑥 𝑑𝑣 = Integrate 𝑢 𝑑𝑣 = 𝑢𝑣 − Exercise 1 𝑥𝑒 𝑥 𝑑𝑥 = 𝑢=𝑥 𝑣 = 𝑒𝑥 New 𝑑𝑢 = 𝑑𝑥 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 Original 𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥 𝑑𝑥 𝑥𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝐶 Exercise 2 Find the average value of 𝑦 = ln 𝑥 on the closed interval 1, 𝑒 . Exercise 3 1 arcsin 𝑥 𝑑𝑥 0 = Exercise 4 Find the particular solution of given the initial condition 𝑑𝑦 𝑑𝑥 𝜋 ,0 2 . = 𝑥 sin 𝑥 Exercise 5 𝑥 2 sin 𝑥 𝑑𝑥 = Exercise 6 sec 3 𝑥 𝑑𝑥 = Exercise 7: AP FRQ Let 𝑓 be the function defined for 𝑥 > 0, with 𝑓 𝑒 = 2 and 𝑓′, the first derivative of 𝑓, given by 𝑓′ 𝑥 = 𝑥 2 ln 𝑥. a) Write an equation for the line tangent to the graph of 𝑓 at the point 𝑒, 2 . b) Is the graph of 𝑓 concave up or concave down on the interval 1 < 𝑥 < 3? Give a reason for your answer. c) Use antidifferentiation to find 𝑓 𝑥 . 8-2: Integration By Parts Objectives: 1. To find an antiderivative using integration by parts Assignment: • P. 531-533: 9-17 odd, 23-33 odd, 39, 43, 49, 53, 73-78, 99, 103a-c, 107 • Homework Supplement