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

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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, $x=4, t=72$ .

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

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

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

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

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

$=8xx5184$ .

$rArr I=41472$ .

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