Question

Let f(x) be a continuous function such that the area bounded by the curve y = f(x), x-axis and the lines x =0 and x = a is ${\displaystyle \frac{a^2}{2}}+{\displaystyle \frac{a}{2}}\sin a+{\displaystyle \frac{\pi }{2}}\cos a$ then $f\left({\displaystyle \frac{\pi }{2}}\right)$
A . 1
B. ${\displaystyle \frac{1}{2}}$
C. ${\displaystyle \frac{1}{3}}$
D. None of these

Answer 1

Hint: In this question, we are given that f(x) be a continuous function and the area bounded by y = f(x) , x-axis and the lines x = 0 and x = a is ${\displaystyle \frac{a^2}{2}}+{\displaystyle \frac{a}{2}}\sin a+{\displaystyle \frac{\pi }{2}}\cos a$. We have to find the value of $f\left({\displaystyle \frac{\pi }{2}}\right)$. For this we differentiate the given equation from a to 0. Then we substitute $a={\displaystyle \frac{\pi }{2}}$ in the equation and get the desirable value of $f\left({\displaystyle \frac{\pi }{2}}\right)$.

Formula Used: Formula for area under curve is = ${\int }_a^{b}f(x)dx$ where (b > a)
We use the formula to differentiate the given equation:-
${\int }_a^{b}f(x)dx$${\displaystyle \frac{d}{dx}}(fg)=f{g}^{\prime }+g{f}^{\prime }$

Complete step by step Solution:
Given that f(x) is a continuous function.
A function that does not have any discontinuity and arbitrarily small changes by restricting enough small changes is called the continuous function.
Area bounded by the lines x = 0 and x = a and y = f(x) will be
= ${\int }_a^{b}f(x)dx$ where (b > a)
And ${\int }_0^{a}f(x)dx={\displaystyle \frac{a^2}{2}}+{\displaystyle \frac{a}{2}}\sin a+{\displaystyle \frac{\pi }{2}}\cos a$
First, we differentiate the above equation w.r.t a, and we get
$f(a)-f(0)={\displaystyle \frac{2a}{2}}+{\displaystyle \frac{1}{2}}((1)\sin a+a\cos a)+{\displaystyle \frac{\pi }{2}}(-\sin a)$
$f(a)=a+{\displaystyle \frac{1}{2}}(\sin a+a\cos a)-{\displaystyle \frac{\pi }{2}}\sin a$
Now we put $a={\displaystyle \frac{\pi }{2}}$, then the equation becomes
$f\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\pi }{2}}$
Then $f\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{1}{2}}$

Therefore, the correct option is (B).

Note: Area under the curve between two points is finding out by doing integral between two points. To find the area under the curve y = f(x) between x = a and x = b, we integrate y = f(x) between the points a and b.