Let f : R rightarrow R be a differentiable fu | vedclass

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(D) Let $L = \lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$.Since $f(1) = 4$,as $x \rightarrow 1$,the integral becomes $\int_{4}^{4} \

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$8f'(1)$

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(D) Let $L = \lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$.Since $f(1) = 4$,as $x \rightarrow 1$,the integral becomes $\int_{4}^{4} \frac{2t \, dt}{0}$,which is an indeterminate form of type $\frac{0}{0}$.We apply $L$'$H$ôpital's rule and the Leibniz integral rule:$L = \lim_{x \rightarrow 1} \frac{\frac{d}{dx} \int_{4}^{f(x)} 2t \, dt}{\frac{d}{dx} (x - 1)}$Using the Leibniz rule,$\frac{d}{dx} \int_{4}^{f(x)} 2t \, dt = 2f(x) \cdot f'(x)$.So,$L = \lim_{x \rightarrow 1} \frac{2f(x) \cdot f'(x)}{1}$.Substituting $x = 1$,we get $L = 2f(1) \cdot f'(1)$.Given $f(1) = 4$,we have $L = 2(4) \cdot f'(1) = 8f'(1)$.

Let $f : R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then the value of $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ is:

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