How do you integrate #int xsqrt(4-x)# using substitution?

1 Answer
George C.
Nov 3, 2016

#int color(white)(.)xsqrt(4-x)color(white)(.) dx = -2/15(3x+8)(4-x)^(3/2) + C#

Explanation:

Let #u = 4-x#

Then #du = -dx#

#int color(white)(.)xsqrt(4-x)color(white)(.) dx = int color(white)(.)(u-4)sqrt(u)color(white)(.) du#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = int color(white)(.)u^(3/2) - 4 u^(1/2)color(white)(.) du#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/5 u^(5/2) - 8/3 u^(3/2) + C#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/5 (4-x)^(5/2) - 8/3 (4-x)^(3/2) + C#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = (2/5 (4-x) - 8/3)(4-x)^(3/2) + C#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/15(3 (4-x) - 20)(4-x)^(3/2) + C#

#color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = -2/15(3x+8)(4-x)^(3/2) + C#

Related questions
See all questions in Integration by Substitution
Impact of this question
7457 views around the world
You can reuse this answer
Creative Commons License