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 two ordinates $$x=0$$ and $$x=a$$ is $$\frac{{{a^2}}}{2} + \frac{a}{2}\sin \,a + \frac{\pi }{2}\cos \,a.$$ Then $$f\left( {\frac{\pi }{2}} \right)$$ is :
A.
$$\frac{1}{2}$$
B.
$$\frac{{{\pi ^2}}}{8} + \frac{\pi }{4}$$
C.
$$\frac{{\pi + 1}}{2}$$
D.
$$\frac{\pi }{2}$$
Answer :
$$\frac{1}{2}$$
Solution :
$$\eqalign{
& {\text{Here, }}\int_0^a {f\left( x \right)dx} = \frac{{{a^2}}}{2} + \frac{a}{2}\sin \,a + \frac{\pi }{2}\cos \,a.\, \cr
& {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}a, \cr
& f\left( a \right) = a + \frac{1}{2}\left( {\sin \,a + a\cos \,a} \right) - \frac{\pi }{2}\sin a \cr
& {\text{So, }}f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + \frac{1}{2}\left( {1 + 0} \right) - \frac{\pi }{2} = \frac{1}{2} \cr} $$