Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x),\ x$-axis and the ordinates $x = \frac{\pi}{4}\$and $x = \beta \frac{\pi}{4}is\pi$ $\left( \beta\sin{\beta + \frac{\pi}{4}\cos{\beta + \sqrt{2}}\beta} \right)\ the\ f\left( \frac{\pi}{2} \right)s$

Question

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x),\ x$-axis and the ordinates $x = \frac{\pi}{4}\$and $x = \beta > \frac{\pi}{4}is\pi$ $\left( \beta\sin{\beta + \frac{\pi}{4}\cos{\beta + \sqrt{2}}\beta} \right)\ the\ f\left( \frac{\pi}{2} \right)s$

Dr. Abubacker · Accepted Answer

(a)
$Given,\ \int_{\pi/4}^{\beta}{f(x)dx = \beta\sin{\beta + \frac{\pi}{4}\cos{\beta + \sqrt{2}}}}\beta$
On differentiating w.r.t. $\beta$ on both sides, we get
$f(\beta) = \sin{\beta + \beta\cos{\beta - \frac{\pi}{4}\sin{\beta + \sqrt{2}}}}$
$Put\ \ \ \ \beta = \frac{\pi}{2}$
$Then,\ f\left( \frac{\pi}{2} \right) = \sin{\frac{\pi}{2} + \frac{\pi}{2}\cos{\frac{\pi}{2} - \frac{\pi}{4}\sin{\frac{\pi}{2} + \sqrt{2} = 1 - \frac{\pi}{4} + \sqrt{2}}}}$

Q. Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x),\ x$-axis and the ordinates $x = \frac{\pi}{4}\$and $x = \beta > \frac{\pi}{4}is\pi$ $\left( \beta\sin{\beta + \frac{\pi}{4}\cos{\beta + \sqrt{2}}\beta} \right)\ the\ f\left( \frac{\pi}{2} \right)s$

Explanation

Reviewed by Dr. Abubacker

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