Indefinite Integration 1 Question 23
24. The value of
$\int _{-2}^{0}\left[x^{3}+3 x^{2}+3 x+3+(x+1) \cos (x+1)\right] d x$ is $(2005,1 M)$
(a) 0
(b) 3
(c) 4
(d) 1
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Solution:
- Let $I=\int _{-2}^{0}\left[x^{3}+3 x^{2}+3 x+3+(x+1) \cos (x+1)\right] d x$
$$ =\int _{-2}^{0}\left[(x+1)^{3}+2+(x+1) \cos (x+1)\right] d x $$
Put $x+1=t$
$$ \begin{aligned} & \Rightarrow \quad d x=d t \\ & \therefore \quad I=\int _{-1}^{1}\left(t^{3}+2+t \cos t\right) d t \\ & =\int _{-1}^{1} t^{3} d t+2 \int _{-1}^{1} d t+\int _{-1}^{1} t \cos t d t \\ & =0+2 \cdot 2[x] _0^{1}+0 \\ & =4 \end{aligned} $$
[since, $t^{3}$ and $t \cos t$ are odd functions]
