Explanation Given: ∫ 0 1 x 2 e 4 x d x Key Concept and Formula Used: By parts ∫ u . v d x = u ∫ v d x − ∫ ( d u d x ∫ v d x ) d x + C where u is first function and v is second function. To decide first and second function we use ILATE where I=Inverse function, L=Logarithmic function, A=Algebraic function, T=Trigonometric function, and E=Exponential function And using ∫ x n d x = x n + 1 n + 1 + C , ∫ e a x d x = e a x a + C , d d x x n = n x n − 1 Calculation: ∫ 0 1 x 2 e 4 x d x we first solve without using limit and then put the limit after integrating. ∫ x 2 e 4 x d x using by parts formula ∫ u . v d x = u ∫ v d x − ∫ ( d u d x ∫ v d x ) d x According to ILATE u = x 2 & v = e 4 x ,putting the value in formula we get ∫ x 2 e 4 x d x = x 2 ∫ e 4 x d x − ∫ ( d d x x 2 ∫ e 4 x d x ) d x = x 2 e 4 x 4 − ∫ ( 2 x × e 4 x 4 ) d x = x 2 e 4 x 4 − 1 2 ∫ ( x e 4 x ) d x (Using ∫ e a x d x = e a x a + C , d d x x n = n x n − 1 ) Again using by parts, according to ILATE u = x & v = e 4 x , = x 2 e 4 x 4 − 1 2 [ x ∫ e 4 x d x − ∫ ( d d x x ∫ e 4 x d x ) d x ] = x 2 e 4 x 4 − 1 2 [ x e 4 x 4 − ∫ ( 1...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8, Problem 13CRE

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$\int_{0}^{1} x^{2} e^{4 x} d x$

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Chapter 8, Problem 12CRE

Chapter 8, Problem 14CRE

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Chapter 8 Solutions

Calculus - Standalone book

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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∫ 0 1 x 2 e 4 x d x

Solution Summary: The author explains how to determine first and second functions using ILATE.

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∫01x2e4xdx

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Chapter 8 Solutions

$\int_{0}^{1} x^{2} e^{4 x} d x$