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(B) Given that the area bounded by the curve $y = f(x)$ from $x = \frac{\pi}{4}$ to $x = \beta$ is $\int_{\pi/4}^{\beta} f(x) dx = \beta \sin \beta +

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that the area bounded by the curve $y = f(x)$ from $x = \frac{\pi}{4}$ to $x = \beta$ is $\int_{\pi/4}^{\beta} f(x) dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta$. Applying the Fundamental Theorem of Calculus,we differentiate both sides with respect to $\beta$: $\frac{d}{d\beta} \left( \int_{\pi/4}^{\beta} f(x) dx \right) = \frac{d}{d\beta} \left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Using the product rule for $\beta \sin \beta$: $f(\beta) = (1 \cdot \sin \beta + \beta \cos \beta) - \frac{\pi}{4} \sin \beta + \sqrt{2}$. Now,substitute $\beta = \frac{\pi}{2}$ into the expression for $f(\beta)$: $f\left( \frac{\pi}{2} \right) = \sin\left( \frac{\pi}{2} \right) + \frac{\pi}{2} \cos\left( \frac{\pi}{2} \right) - \frac{\pi}{4} \sin\left( \frac{\pi}{2} \right) + \sqrt{2}$. Since $\sin\left( \frac{\pi}{2} \right) = 1$ and $\cos\left( \frac{\pi}{2} \right) = 0$: $f\left( \frac{\pi}{2} \right) = 1 + 0 - \frac{\pi}{4}(1) + \sqrt{2} = 1 - \frac{\pi}{4} + \sqrt{2}$.

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