MCQ
Let $f:R \to R$ be a differentiable function having $f(2) = 6,f'(2) = \left( {\frac{1}{{48}}} \right).$ Then $\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f(x)} {\frac{{4{t^3}}}{{x - 2}}} dt$ equals
- A$12$
- ✓$18$
- C$24$
- D$36$
$ = 4{(f(2))^3} \times f'(2) = 18$.
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