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How do you evaluate #int (x^2(x^3 + 1)^3)# for [0, 1]?

1 Answer

Oct 15, 2015

#5/4#

Explanation:

#x^3+1=t => 3x^2dx=dt => x^2dx= dt/3#

#t_1 = 0^3+1 = 1#
#t_2 = 1^3+1 = 2#

#I = int_0^1 x^2(x^3+1)^3dx = int_1^2 t^3 dt/3 = 1/3 t^4/4= 1/12 t^4#

#I = 1/12 (2^4-1^4) = 1/12 (16-1) = 15/12 = 5/4#

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