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

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(A) Let $L = \lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$. Evaluating the integral, we get: $L = \lim _{x \rightarrow 1} \frac{1}{x-1} [t

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let $L = \lim _{x ightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$. Evaluating the integral, we get: $L = \lim _{x ightarrow 1} \frac{1}{x-1} [t^2]_4^{f(x)} = \lim _{x ightarrow 1} \frac{[f(x)]^2 - 16}{x-1}$. Since $f(1) = 4$, the expression is in the $\frac{0}{0}$ form. Applying $L'H\hat{o}pital's$ rule: $L = \lim _{x ightarrow 1} \frac{\frac{d}{dx} ([f(x)]^2 - 16)}{\frac{d}{dx} (x-1)} = \lim _{x ightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}$. Substituting the values $f(1) = 4$ and $f^{\prime}(1) = 2$: $L = 2 	imes f(1) 	imes f^{\prime}(1) = 2 	imes 4 	imes 2 = 16$.

Let $f: R \rightarrow R$ be a differentiable function and $f(1)=4$. Then the value of $\lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$, if $f^{\prime}(1)=2$ is

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