If $f(x)$ be continuous function such that the area bounded by the curve $y = f(x)$, the $x$-axis and the lines $x = a\ and\ x = 0\ is\frac{a^{2}}{2} + \frac{a}{2}\sin{a + \frac{\pi}{2}\cos a.}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Value\ of\ f\left( \frac{\pi}{2} \right)\ is$

Question

Dr. Abubacker · Accepted Answer

(a)
According to the given condition,
$Area\ of\ curve = \int_{0}^{a}{f(x)dx}$
$\Longrightarrow \frac{a^{2}}{2} + \frac{a}{2}\sin{a + \frac{\pi}{2}\cos a}$
$= \int_{0}^{a}{f(x)dx}$
On differentiating both sides w.r.t. $a$, we get
$a + \frac{1}{2}\sin{a + \frac{a}{2}\cos{a - \frac{\pi}{2}\sin{a = f(a)}}}$
$\Longrightarrow \ \ f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} + \frac{1}{2}\sin{\frac{\pi}{2} + \frac{\pi}{4}\cos{\frac{\pi}{2} - \frac{\pi}{2}\sin\frac{\pi}{2}}}$
$\Longrightarrow \ \ f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} + \frac{1}{2} - \frac{\pi}{2}$
$\Longrightarrow \ \ f\left( \frac{\pi}{2} \right) = \frac{1}{2}$

Q. If $f(x)$ be continuous function such that the area bounded by the curve $y = f(x)$, the $x$-axis and the lines $x = a\ and\ x = 0\ is\frac{a^{2}}{2} + \frac{a}{2}\sin{a + \frac{\pi}{2}\cos a.}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Value\ of\ f\left( \frac{\pi}{2} \right)\ is$

Explanation

Reviewed by Dr. Abubacker

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