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$\int_{-2}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x$
is equals to
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The correct answer is:
$4$
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.
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.
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