How do you integrate #int x^2(x^3+8)^2dx# from [-2,4]?

Question

1982 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-24T16:54:20

#41472#

the Integrand is given by
#x^2(x^6+16x^3+64)=x^8+16x^5+64x^2#
and a primitive is #x^9/9+16/6x^6+64/3x^3#

Answer 2 · 2018-05-24T18:16:38

# 41472#.

Suppose that, #I=int_-2^4x^2(x^3+8)^2dx#.

Let, #(x^3+8)=t. :. 3x^2dx=dt, or, x^2dx=1/3dt#.

Also, when #x=-2, t=x^3+8=(-2)^3+8=0# and,

similarly when, #x=4, t=72#.

#:. I=int_-2^4(x^3+8)^2x^2dx#,

#=int_0^72t^2*1/3dt#,

#=1/3[1/3t^3]_0^72#,

#=1/9[72^3-0]#,

#=1/9*72*72^2#,

#=8xx5184#.

#rArr I=41472#.