Explanation Given Information: The provided integral is ∫ [ ( 2 x − 1 ) e 2 x 2 − 2 x + x e x 2 ] d x . Formula used: Follow the steps as mentioned below to solve an integral ∫ f d x by using substitution rule. Step 1 : Assume a function u in terms of x from f . Step 2 : Take the derivative d u d x of the function obtained in step 1 and solve for d x in terms of d u . Step 3 : Substitute the value of u from step 1 and the value of d x from step 2 in the provided integral and then solve the integral. The sum for an integral, ∫ [ f ( x ) + g ( x ) ] d x = ∫ f ( x ) d x + ∫ g ( x ) d x Where f ( x ) and g ( x ) are two functions. The constant multiple rule, ∫ k ⋅ f ( x ) d x = k ∫ f ( x ) d x Where k is a constant. The rule for the indefinite integral of the exponential e x , ∫ e x d x = e x + C Where C is the constant of integration. Calculation: Consider the integral, ∫ [ ( 2 x − 1 ) e 2 x 2 − 2 x + x e x 2 ] d x To evaluate the integral ∫ [ ( 2 x − 1 ) e 2 x 2 − 2 x + x e x 2 ] d x apply sum for an integral. ∫ [ ( 2 x − 1 ) e 2 x 2 − 2 x + x e x 2 ] d x = ∫ ( 2 x − 1 ) e 2 x 2 − 2 x d x + ∫ x e x 2 d x Integrate individual terms of the expression ∫ ( 2 x − 1 ) e 2 x 2 − 2 x d x + ∫ x e x 2 d x and then combine the results. Integrate first term ∫ ( 2 x − 1 ) e 2 x 2 − 2 x d x by using substitution rule as, Assume a function u in terms of x from ( 2 x − 1 ) e 2 x 2 − 2 x , u = 2 x 2 − 2 x Take the derivative d u d x of the function u = 2 x 2 − 2 x , d u d x = 4 x − 2 = 2 ( 2 x − 1 ) Solve for d x , d x = 1 2 ( 2 x − 1 ) d u Now substitute u = 2 x 2 − 2 x and d x = 1 2 ( 2 x − 1 ) d u in the integral term ∫ ( 2 x − 1 ) e 2 x 2 − 2 x d x , this gives, ∫ ( 2 x − 1 ) e 2 x 2 − 2 x d x = ∫ ( 2 x − 1 ) ⋅ e u ⋅ 1 2 ( 2 x − 1 ) d u = ∫ ( 2 x − 1 ) 2 ⋅ e u ⋅ 1 ( 2 x − 1 ) d u Simplify the integral by cancelling 2 x − 1 by 2 x − 1 . ∫ ( 2 x − 1 ) 2 ⋅ e u ⋅ 1 ( 2 x − 1 ) d u = ∫ e u 2 d u Evaluate the integral ∫ e u 2 d u by using constant multiple rule and rule for the indefinite integral of the exponential e x . ∫ e u 2 d u = 1 2 ∫ e u d u = 1 2 e u + C Therefore, the solution of the integral ∫ 1 2 e u d u is 1 2 e u + C

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

7th Edition

ISBN: 9781337291293

Author: Waner, Stefan; Costenoble, Steven

Publisher: Brooks Cole

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Chapter 6.2, Problem 51E

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To calculate: The solution of the integral ∫[(2x−1)e2x2−2x+xex2]dx.

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Chapter 6.2, Problem 50E

Chapter 6.2, Problem 52E

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Chapter 6 Solutions

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

Ch. 6.1 - In Exercises 142, evaluate the integral. [HinT:...Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - In Exercises 142, evaluate the integral. [HinT:...Ch. 6.1 - In Exercises 142, evaluate the integral. [HinT:...Ch. 6.1 - In Exercises 142, evaluate the integral. [HinT:...Ch. 6.1 - Prob. 10E

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The solution of the integral ∫ [ ( 2 x − 1 ) e 2 x 2 − 2 x + x e x 2 ] d x .

Solution Summary: The author explains how to solve an integral using substitution rule.

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To calculate: The solution of the integral ∫[(2x−1)e2x2−2x+xex2]dx.

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Chapter 6 Solutions