Question
Let $$f\left( x \right)$$ be a continuous function such that the area bounded by the curve $$y = f\left( x \right),$$ the $$x$$-axis, and the lines $$x=0$$ and $$x=a$$ is $$1 + \frac{{{a^2}}}{2}\sin \,a.$$ Then :
A.
$$f\left( {\frac{\pi }{2}} \right) = 1 + \frac{{{\pi ^2}}}{8}$$
B.
$$f\left( a \right) = 1 + \frac{{{a^2}}}{2}\sin \,a$$
C.
$$f\left( a \right) = a\sin \,a + \frac{1}{2}{a^2}\cos \,a$$
D.
none of these
Answer :
$$f\left( a \right) = a\sin \,a + \frac{1}{2}{a^2}\cos \,a$$
Solution :
$$\eqalign{
& {\text{Here }}\int_0^a {f\left( x \right)dx} = 1 + \frac{{{a^2}}}{2}\sin \,a.\,\,{\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}a, \cr
& f\left( a \right) = a\sin \,a + \frac{{{a^2}}}{2}\cos \,a \cr} $$