Question
If \[f\left( x \right) = \left\{ \begin{array}{l}
{e^{\cos \,x}}\sin \,x,\,{\rm{for }}\left| x \right| \le 2\\
2,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}}
\end{array} \right.,\] then $$\int\limits_{ - 2}^3 {f\left( x \right)dx} = ?$$
A.
$$0$$
B.
$$1$$
C.
$$2$$
D.
$$3$$
Answer :
$$2$$
Solution :
\[\begin{array}{l}
{\rm{If }}f\left( x \right) = \left\{ \begin{array}{l}
{e^{\cos \,x}}\sin \,x,\,{\rm{for }}\left| x \right| \le 2\\
2,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}}
\end{array} \right\}\\
= \left\{ \begin{array}{l}
{e^{\cos \,x}}\sin \,x,\,{\rm{for }} - 2 \le x \le 2\\
2,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}}
\end{array} \right\}
\end{array}\]
$$\eqalign{
& \int\limits_{ - 2}^3 {f\left( x \right)} dx = \int\limits_{ - 2}^2 {f\left( x \right)} dx + \int\limits_2^3 {f\left( x \right)} dx \cr
& = \int\limits_{ - 2}^2 {{e^{\cos \,x}}\sin \,x\,dx} + \int\limits_2^3 {2\,dx} = 0 + 2\left[ x \right]_2^3 \cr
& \left[ {\because {e^{\cos \,x}}\sin \,x\,{\text{ is an odd function}}{\text{.}}} \right] \cr
& = 2\left[ {3 - 2} \right] = 2 \cr
& \therefore \int\limits_{ - 2}^3 {f\left( x \right)} dx = 2 \cr} $$