Question

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

• A. $-4$
• B. $-3$
• C. $-2$
• D. $-1$

Answer 1

The correct option is C $-2$

We can write
$I={\int }_{-2}^0\left[\right(x+1{)}^3-1+(x+1)\cos (x+1)]dx,$
Substitute $x+1=t,$ so that
$I={\int }_{-1}^1[t^3-1+t\cos t]dt,$
$I={\int }_{-1}^1(-1)dt=t{]}_{-1}^1=-2$
as $t^3+t\cos t$ is an odd function.