Let f(x) be a non-negative continuous functio | vedclass

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(D) Given,$\int_{\pi/4}^{\beta} f(x) dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta$. On differentiating both sides with respect to

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

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(D) Given,$\int_{\pi/4}^{\beta} f(x) dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta$. On differentiating both sides with respect to $\beta$ using the Leibniz integral rule,we get: $f(\beta) = \frac{d}{d\beta} \left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Applying the product rule and derivative rules: $f(\beta) = (1 \cdot \sin \beta + \beta \cos \beta) + \frac{\pi}{4} (-\sin \beta) + \sqrt{2}$. $f(\beta) = \sin \beta + \beta \cos \beta - \frac{\pi}{4} \sin \beta + \sqrt{2}$. To find $f\left( \frac{\pi}{2} \right)$,substitute $\beta = \frac{\pi}{2}$: $f\left( \frac{\pi}{2} \right) = \sin \left( \frac{\pi}{2} \right) + \frac{\pi}{2} \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{4} \sin \left( \frac{\pi}{2} \right) + \sqrt{2}$. Since $\sin \left( \frac{\pi}{2} \right) = 1$ and $\cos \left( \frac{\pi}{2} \right) = 0$: $f\left( \frac{\pi}{2} \right) = 1 + \frac{\pi}{2}(0) - \frac{\pi}{4}(1) + \sqrt{2}$. $f\left( \frac{\pi}{2} \right) = 1 - \frac{\pi}{4} + \sqrt{2}$.

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 $\left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Then $f\left( \frac{\pi}{2} \right)$ is

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