Question

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

Answer 1

${\int }_{-2}^0{x}^{3}dx=[\frac{x^4}{4}{]}_{-2}^0=-4$

Similarily for

$3x^2=8$

$3x=-6$

And for $(x+1)\cos (x+1)$, we have

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

${\int }_{-1}^{1}m\cos mdm$

This is equal to zero as it is an odd function.

Hence, Value =$-6-4+8=-2$