Q111 / 180 · KCET 2023 · Mathematics

$\int_{-2}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x$ is equals to

  • A
    $3$
  • Correct answer
    B
    $4$
  • C
    $1$
  • D
    $0$

Solution

Let $\begin{aligned} I & =\int_{-2}^0 x^3+3 x^2+3 x+3+(x+1) \cos (x+1) d x \\ I & =\int_{-2}^0\left[(x+1)^3+2+(x+1) \cos (x+1) d x\right.\end{aligned}$ Putting, $x+1=t$. Then $d x=d t$, we get $\begin{aligned} & I=\int_{-1}^1\left[t^3+2+t \cos t\right] d t \\ & I=\int_{-1}^1(2) \text { u. }=[2 t]_{-1}^1=2[1-(-1)]=4\end{aligned}$ As $t^3+t \cos t$ is an odd function.

Hence, the correct answer is option: 'B'.

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