Let $f: R \to R$ be a differentiable function having $f(2) =6, f'(2) =(\frac{1}{48})$. Find $\lim_{x\to 2} \int^{f(x)}_6 \frac{4t^3}{x-2} \, dt$ Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(2) =6, f'(2) =(\frac{1}{48})$. Find $\lim_{x\to 2} \int^{f(x)}_6 \frac{4t^3}{x-2}\,dt$.
Please suggest how to proceed in such questions... will be of great help.. thanks. 
 A: Note that $$\int _6^{f(x)}4t^3 =(f(x))^4-6^4$$
Thus $$\lim_{x\to 2} \frac {1}{x-2} \int _6^{f(x)}4t^3=$$
$$\lim_{x\to 2} \frac {(f(x))^4-6^4}{x-2} $$
Therefore,  $$\lim_{x\to 2} \frac {(f(x))^4-6^4}{x-2}= 4(f(x))^3(f'(x))=18$$
A: Alternatively you could do L'Hopital rule first and avoid integration altogether:
$$\lim_{x\to 2} \frac{\int_6^{f(x)} 4t^3 dt}{x-2} = \lim_{x\to 2} \frac{4(f(x))^3f'(x)}{1} = 18$$
This approach is more useful since it allows us to compute the limit even if the integral is not doable by hand, or could save time computing a nasty integral.
A: You have
$$
\int^{f(x)}_6 4t^3\cdot\frac 1 {x-2} \, dt. \tag 1
$$
First note that as $t$ goes from $6$ to $f(x),$ the factor $\dfrac 1 {x-2}$ does not change. Therefore $(1)$ is equal to
$$
\left(\int_6^{f(x)} 4t^3 \, dt\right) \cdot \frac 1 {x-2}, \tag 2
$$
i.e. the factor $\dfrac 1 {x-2}$ is entirely outside of the integral. In this case you could easily evaluate the integral, but you don't actually need that.
Observe that $(2)$ is equal to
$$
\int_{f(2)}^{f(x)} 4t^2\,dt \cdot \frac 1 {x-2}. \tag 3
$$
Let $\displaystyle g(x) = \int_{f(2)}^{f(x)} 4t^3\,dt.$ Then $(3)$ becomes
$$
\big( g(x) - g(2)\big) \cdot \frac 1 {x-2}
$$
or
$$
\frac{g(x)-g(2)}{x-2}.
$$
The definition of differentiation then tells us that
$$
\lim_{x\to2} \frac{g(x)-g(2)}{x-2} = g'(2).
$$
So we're looking for $g'(2).$
We have
$$
y = g(x) = \int_6^u 4t^3\cdot dt \quad \text{and } u = f(x).
$$
So
$$
\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = 4u^3\cdot f'(x).
$$
Now plug in $x=2,$ which makes $u=6.$
