Question

Evaluate:${\int }_{-2}^0(x^3+3x^2+3x+3+(x+1)\cos (x+1))dx$

Answer 1

${\int }_{-2}^0(x^3+3x^2+3x+3+(x+1)\cos (x+1))dx$

Let $x+1=t$

When $x=-2\Rightarrow t=-1$

When $x=0\Rightarrow t=1$

$\Rightarrow I={\int }_{-1}^1(t^3+2+t\cos t)dt$

$={\int }_{-1}^{1}2dt$ [since $t^3+t\cos t$ is an odd function]

$={\left[2t\right]}_{-1}^1$

$=2[1+1]=4$