Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_{6}^{f(x)} \frac{4 t^{3}}{x-2} d t$ equals

JEE Main · 2005 · 08 May 2005 · Offline · Definite Integrals

Let

f : R \to R

be a differentiable function having

f (2) = 6, f^{'} (2) = (\frac{1}{48})

. Then

lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t

equals

Answer: (D)

x \to 2

both the integral

\int_{6}^{f (x)} 4 t^{3} d t

and the factor

x - 2

tend to

0

, so the quotient has the indeterminate form

\frac{0}{0}

. Differentiating numerator and denominator with respect to

x

, the numerator's derivative by the Leibniz rule is

4 f (x)^{3} f^{'} (x)

and the denominator's is

1

. The limit is therefore

4 f (2)^{3} f^{'} (2) = 4 \cdot 6^{3} \cdot \frac{1}{48} = \frac{4 \cdot 216}{48} = 18

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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