How do you integrate #int_-1^1x(1+x)^3dx#?

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Answer 1 · 2017-03-26T20:04:03

Do a "u" substitute

Given: #int_-1^1x(1+x)^3dx#

let #u = x + 1#, then #du = dx and x = u -1#

Change the limits:

#a = -1 + 1#

#a = 0#

#b = 1 + 1#

#b = 2#

#int_-1^1x(1+x)^3dx = int_0^2(u-1)u^3du#

#int_-1^1x(1+x)^3dx = int_0^2u^4-u^3du#

#int_-1^1x(1+x)^3dx = u^5/5-u^4/4|_0^2#

#int_-1^1x(1+x)^3dx = 2^5/5-2^4/4#

#int_-1^1x(1+x)^3dx = 12/5#