For this problem, fill in each blank with the correct answer. Syntax example: to enter r? sin (), type x^2sin(x/2) (do not use any space or * for multiplication and use parentheses for the input/argument of the sine function). To enter x? (- sin(:), type x^2(- sin(x/2)) (use parentheses when multiplying by a negative and do not use any space or * for multiplication) or just simplify to -x^2sin(x/2). It is always best to simplify your answers as much as possible. (For instance, if the answer is 2/(1/2), simplify to 4, if the answer is 2(-3) simplify to -6, if the answer is 2x(-x^2) simplify to -2x^3, etc.) Note: DO NOTI include the "dx" part in your answers below! It's already there. Evaluate the indefinite integralf x? sin ()da by following the steps outlined below. We will use the method of integration by parts which is derived from the for differentiation. We first apply integration by parts with , dv = dx du = dx, v = to get: Ja? sin(플) dz= +4/zcos (특) da. Next, we apply integration by parts to the integralf xcos () da with , dv = dx du = dr, v = to finally obtain: Ju" sin(플) dz= 1-/2sin(들)da) +4( +8 +16 cos(특) + C.
For this problem, fill in each blank with the correct answer. Syntax example: to enter r? sin (), type x^2sin(x/2) (do not use any space or * for multiplication and use parentheses for the input/argument of the sine function). To enter x? (- sin(:), type x^2(- sin(x/2)) (use parentheses when multiplying by a negative and do not use any space or * for multiplication) or just simplify to -x^2sin(x/2). It is always best to simplify your answers as much as possible. (For instance, if the answer is 2/(1/2), simplify to 4, if the answer is 2(-3) simplify to -6, if the answer is 2x(-x^2) simplify to -2x^3, etc.) Note: DO NOTI include the "dx" part in your answers below! It's already there. Evaluate the indefinite integralf x? sin ()da by following the steps outlined below. We will use the method of integration by parts which is derived from the for differentiation. We first apply integration by parts with , dv = dx du = dx, v = to get: Ja? sin(플) dz= +4/zcos (특) da. Next, we apply integration by parts to the integralf xcos () da with , dv = dx du = dr, v = to finally obtain: Ju" sin(플) dz= 1-/2sin(들)da) +4( +8 +16 cos(특) + C.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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