Explanation Given: • ∫ x 4 ⋅ e 7 x ⋅ d x • Reduction Formula: ∫ x n ⋅ e x ⋅ d x = x n ⋅ e x − n ∫ x n − 1 ⋅ e x ⋅ d x Key Concepts Used: Many trigonometric functions can be integrated by combining substitution and Integration by parts with the appropriate trigonometric identities . Here in the given problem we have to find ∫ x 4 ⋅ e 7 x ⋅ d x that means we have to find the integral value of x 4 ⋅ e 7 x . We have to use the reduction formula ∫ x n ⋅ e x ⋅ d x = x n ⋅ e x − n ∫ x n − 1 ⋅ e x ⋅ d x to find the given integral ∫ x 4 ⋅ e 7 x ⋅ d x . Here instead of e x , we got e 7 x . Thereby we have to go with substitution process. Here we substitute u = 7 x and d u = 7 ⋅ d x . Thereby x should become as u 7 in the given integral. Thereby the given integral is in terms of u alone. Now we have to do the further integration by using the reduction formula ∫ x n ⋅ e x ⋅ d x = x n ⋅ e x − n ∫ x n − 1 ⋅ e x ⋅ d x . Then in the final step we have to replace the dummy variable by the original variable. Formula Used: Reduction Formula: ∫ x n ⋅ e x ⋅ d x = x n ⋅ e x − n ∫ x n − 1 ⋅ e x ⋅ d x . Calculation: Description: ∫ x n ⋅ e x ⋅ d x = x n ⋅ e x − n ∫ x n − 1 ⋅ e x ⋅ d x Step 1 : Write the given integral. ∫ x 4 ⋅ e 7 x ⋅ d x Step 2 : Here instead of e x , we got e 7 x . Thereby we have to go with substitution process. Here we substitute u = 7 x . ∫ x 4 ⋅ e 7 x ⋅ d x S u b s t i t u t e u = 7 x ⇒ x = u 7 d u d x = d d x ( 7 x ) = 7 ⇒ d u = 7 ⋅ d x T h e r e b y ∫ x 4 ⋅ e 7 x ⋅ d x = ∫ ( u 7 ) 4 ⋅ e u ⋅ d u 7 = 1 7 4 ∫ u 4 ⋅ e u ⋅ d u 7 ∫ x 4 ⋅ e 7 x ⋅ d x = 1 7 5 ∫ u 4 ⋅ e u ⋅ d u Step 3 : Compare the above integral ∫ u 4 ⋅ e u ⋅ d u with the formula, we found that the value of n = 4 . Thereby we substitute n = 4 in the formula. Then again we have to substitute for n = 3 , n = 2 , n = 1 the same reduction formula to find the value of the integral ∫ u 4 ⋅ e u ⋅ d u ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.1, Problem 62E

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The value of the given integral by using substitution and the reduction formula from exercise $60$

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Chapter 8.1, Problem 61E

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Chapter 8 Solutions

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Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The value of the given integral by using substitution and the reduction formula from exercise 60

Solution Summary: The author explains how many trigonometric functions can be integrated by combining substitution and Integration by parts with the appropriate trigonemetric identities.

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The value of the given integral by using substitution and the reduction formula from exercise $60$

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Chapter 8 Solutions