int_9^x frac{f(y)}{y^2} , dy = 2 sqrt{x} - 6 | vedclass

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Question

(B) Given the integral equation: $\int_9^x \frac{f(y)}{y^2} \, dy = 2 \sqrt{x} - 6$. Applying the Leibniz integral rule to differentiate both sides wi

Vedclass · Accepted Answer

$x \sqrt{x}$

Vedclass · Answer

(B) Given the integral equation: $\int_9^x \frac{f(y)}{y^2} \, dy = 2 \sqrt{x} - 6$. Applying the Leibniz integral rule to differentiate both sides with respect to $x$: $\frac{d}{dx} \left( \int_9^x \frac{f(y)}{y^2} \, dy \right) = \frac{d}{dx} (2 \sqrt{x} - 6)$. Using the Fundamental Theorem of Calculus,the derivative of the left side is $\frac{f(x)}{x^2}$. The derivative of the right side is $2 \cdot \frac{1}{2 \sqrt{x}} = \frac{1}{\sqrt{x}}$. Equating both sides: $\frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}$. Solving for $f(x)$: $f(x) = \frac{x^2}{\sqrt{x}} = x^{2 - 1/2} = x^{3/2} = x \sqrt{x}$.

$\int_9^x \frac{f(y)}{y^2} \, dy = 2 \sqrt{x} - 6 \implies f(x) = ?$

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