If f:R to R is a differentiable function and | vedclass

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(A) Let $L = \lim_{x 	o 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$.Since the integrand $\frac{2t}{x-1}$ does not depend on $t$ in the denominator,we can w

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$8f'(1)$

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(A) Let $L = \lim_{x 	o 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$.Since the integrand $\frac{2t}{x-1}$ does not depend on $t$ in the denominator,we can write the integral as $\frac{1}{x-1} \int_4^{f(x)} 2t dt$.The integral of $2t$ is $t^2$,so $\int_4^{f(x)} 2t dt = [t^2]_4^{f(x)} = (f(x))^2 - 16$.Thus,$L = \lim_{x 	o 1} \frac{(f(x))^2 - 16}{x - 1}$.Using the difference of squares,$(f(x))^2 - 16 = (f(x) - 4)(f(x) + 4)$.Since $f(1) = 4$,as $x 	o 1$,$f(x) 	o 4$,so $f(x) + 4 	o 8$.Therefore,$L = \lim_{x 	o 1} \frac{(f(x) - 4)(f(x) + 4)}{x - 1} = \left( \lim_{x 	o 1} \frac{f(x) - f(1)}{x - 1} ight) 	imes \left( \lim_{x 	o 1} (f(x) + 4) ight)$.This simplifies to $f'(1) 	imes (4 + 4) = 8f'(1)$.

If $f:R \to R$ is a differentiable function and $f(1) = 4$,then the value of $\lim_{x \to 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$ is:

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