Question

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

• A. $3$
• B. $0$
• C. $4$
• D. $6$

Answer 1

The correct option is C $4$

Let $I={\int }_{-2}^0[x^3+3x^2+3x+3+(x+1)\cos (x+1)]dx$
$={\int }_{-2}^0[{(x+1)}^3+2+(x+1)\cos (x+1)]dx$
Put $x+1=t\Rightarrow dx=dt$
$\therefore I={\int }_{-1}^1(t^3+2+t\cos t)dt={\int }_{-1}^1{t}^{3}dt+{\int }_{-1}^{1}2dt+{\int }_{-1}^{1}t\cos tdt=0+{\left[2t\right]}_{-1}^1+0=4$