3. Integration By Parts (Short-Cut Method) 3. *Sx* e-2x dxe-2x(2x4 + 4x³ + 6x² + 6x + 3) + C4A2xe(2x4 + 4x3 + 6x² + 6x + 3) + C433°x²e-2x +хе2.3x4 e-2x – x³e-2x .2Ce-2x + C4.-2x2.3хе2.x4 -2x + x3e--2x-2xe2xx2e-2x+ Ce4AO BE None of the Above+HIN

Answered: Image /qna-images/answer/460387fd-3812-4b42-b758-8a5876913af0.jpg

Sx*e-2x dx e-2 (2x* + 4x³ + 6x² + 6x + 3) + C 4 A e-2x (2x* + 4x3 + 6x² + 6x + 3) + C B -2x 3 x* e-2x – x³e-2x -2x + C 4 + xe -2x 2 3 x²e-2x 2. D x* e-2x + x3e-2x + xe-2r e-2x + C E None of the Above 3/2

Sxe-2x dx e-2 (2x + 4x³ + 6x² + 6x + 3) + C 4 A e-2x (2x* + 4x3 + 6x² + 6x + 3) + C B -2x 3 x* e-2x – x³e-2x -2x + C 4 + xe -2x 2 3 x²e-2x 2. D x* e-2x + x3e-2x + xe-2r e-2x + C E None of the Above 3/2

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter6: Topics In Analytic Geometry

Section6.4: Hyperbolas

Problem 5ECP: Repeat Example 5 when microphone A receives the sound 4 seconds before microphone B.

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Question

3. Integration By Parts (Short-Cut Method)

3. *
Sx* e-2x dx
e-2x
(2x4 + 4x³ + 6x² + 6x + 3) + C
4
A
2x
e
(2x4 + 4x3 + 6x² + 6x + 3) + C
4
3
3
°x²e-2x +
хе
2.
3
x4 e-2x – x³e-2x .
2
C
e-2x + C
4.
-2x
2.
3
хе
2.
x4 -2x + x3e-
-2x
-2x
e
2x
x2e-2x
+ C
e
4
A
O B
E None of the Above
+
HIN

With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.

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