Let f: R R be a differentiable function having f(2)=6, f'(2)=148.

Question

Let $f: R 	o R$ be a differentiable function having $f(2)=6$, $f'(2)=\dfrac{1}{48}$. Then $\displaystyle\lim_{x	o2} \int_6^{f(x)} \dfrac{4t^3}{x-2}\,dt$ equals:

Accepted Answer

$\lim_{x	o2} \dfrac{1}{x-2}\int_6^{f(x)} 4t^3\,dt$

Applying L'Hospital's rule (differentiating w.r.t. $x$):

$\lim_{x	o2} \dfrac{4[f(x)]^3 f'(x)}{1} = 4[f(2)]^3f'(2) = 4	imes6^3	imes\dfrac{1}{48} = 4	imes216	imes\dfrac{1}{48} = 18$

Solutions · 1