Question

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

• A. $-4$
• B. $0$
• C. $4$
• D. None of these

Answer 1

The correct option is C $4$

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

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

$={\int }_{-2}^1{[u\cos udu+\frac{{(x+1)}^4}{4}]}_{-2}^0+{\left[2x\right]}_{-2}^0$

$={[(x+1)\sin (x+1)-\cos (x+1)]}_{-2}^0+\frac{1}{4}-\frac{1}{4}+0+4$

$=\sin 1-\cos 1+\sin (-1)+\cos (-1)+4$

$=\sin 1-\cos 1-\sin 1+\cos 1+4$

$=4$

Hence, the answer is $4.$