Question

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

• A. $-4$
• B. $0$
• C. $4$
• D. $6$

Answer 1

The correct option is C $4$

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

Put $x+1=t\Rightarrow dx=dt$

$\therefore I={\int }_{-1}^1(t^3+2+t\cos t)dt={\int }_{-1}^1{t}^{3}dt+2{\int }_{-1}^{1}2dt+{\int }_{-1}^{1}t\cos tdt=0+4{\left[x\right]}_0^1+0=4$