If x cdot sin(pi x) = int_{0}^{x^2} f(t) , dt | vedclass

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(A) Given the equation: $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign with

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) Given the equation: $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign with respect to $x$: $\frac{d}{dx} [x \cdot \sin(\pi x)] = \frac{d}{dx} \int_{0}^{x^2} f(t) \, dt$. Using the product rule on the left side and the Fundamental Theorem of Calculus on the right side: $1 \cdot \sin(\pi x) + x \cdot \cos(\pi x) \cdot \pi = f(x^2) \cdot \frac{d}{dx}(x^2)$. $\sin(\pi x) + \pi x \cos(\pi x) = f(x^2) \cdot 2x$. To find $f(4)$,we set $x^2 = 4$,which implies $x = 2$ or $x = -2$. Case $1$: Let $x = 2$. $\sin(2\pi) + \pi(2) \cos(2\pi) = f(4) \cdot 2(2)$. $0 + 2\pi(1) = 4 \cdot f(4)$. $2\pi = 4 \cdot f(4) \implies f(4) = \frac{\pi}{2}$. Case $2$: Let $x = -2$. $\sin(-2\pi) + \pi(-2) \cos(-2\pi) = f(4) \cdot 2(-2)$. $0 - 2\pi(1) = -4 \cdot f(4)$. $-2\pi = -4 \cdot f(4) \implies f(4) = \frac{\pi}{2}$. Thus,the value of $f(4)$ is $\frac{\pi}{2}$.

If $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$ where $f$ is a continuous function,then the value of $f(4)$ is:

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