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

Question

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

Accepted Answer

$\left( 1-\frac{\pi }{4}+\sqrt{2} \right)$

Answer

$\left( 1-\frac{\pi }{4}-\sqrt{2} \right)$

Answer

$\left( \frac{\pi }{4}+\sqrt{2}-1 \right)$

Answer

$\left( \frac{\pi }{4}-\sqrt{2}+1 \right)$

definite-integral

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

definite-integral

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