Definite Integration Question 23
Question 23
- The value of
$\int_{-2}^{0}\left[x^{3}+3 x^{2}+3 x+3+(x+1) \cos (x+1)\right] d x$ is $(2005,1 \mathrm{M})$ (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]
