Integration with infinity: $\int_4^\infty f\,'(x)e^{-x/4}\, dx$ Let $f$ be a continuous function defined on the interval $[2, \infty[$ such that
$f(4)=14$, $|f(x)| < x^3+10$, 
and $$\int_4^\infty f(x)e^{-x/4} dx=-5\;.$$
Determine the value of:
$$\int_4^\infty f\,'(x)e^{-x/4} dx\;.$$
 A: Hint: Try (definite) integration by parts:
$$\int_a^b f'(x) g(x) dx = \left[f(x) g(x)\right]_a^b - \int_a^b f(x) g'(x) dx.$$
Apply this to a suitable choice of functions $f(x), g(x)$ and bounds $a,b$. You should be able to find the right side of the equation using the conditions on $f$ provided by the question.

Solution: For completeness, let us see what we get. We apply the above to $f(x) = f(x), g(x) = e^{-x/4}, a = 4, b = \infty$. Finally we use e.g. the squeeze theorem to find the limit in the second line below to get the final result.
$$\begin{align}
\int_4^{\infty} f'(x) e^{-x/4} dx &= \left[f(x) e^{-x/4}\right]_4^{\infty} - \int_4^{\infty} f(x) \frac{-1}{4} e^{-x/4} dx \\ 
&= \left[\lim_{x \to \infty} f(x) e^{-x/4} - f(4) e^{-4/4}\right] + \frac{1}{4} \left(\int_4^{\infty} f(x) e^{-x/4} dx\right) \\
&= 0 - \frac{14}{e} - \frac{5}{4}
\end{align}$$
A: Hint: Use integration by parts.
A: Since it is given that 
$$\int_4^\infty f(x)e^{-x/4} dx=-5\;$$
$$
\begin{align*}
f(x)e^{-x/4} dx &= -4 e^{\frac{-x}{4}} f(x) + 4 \int f'(x)e^{\frac{-x}{4}} dx \tag{A}\\
\end{align*}
$$
In order to apply limits, you might have to think what to do next.




 There is a reason why it is given that $|f(x)| < x^3+10 \hspace{5pt}$  does that mean that when you apply the limit as 

as $x \rightarrow \small{\infty}$ 

$$\displaystyle{lim_{x \to \infty } e^{\frac{-x}{4}} f(x)} = {lim_{x \to \infty } \frac{f(x)} {e^{\frac{x}{4}}}}  \rightarrow 0$$

Therefore applying limits in $(A)$, you should get 

$$\frac{4f(14)}{e} + 4 \int f'(x)e^{\frac{-x}{4}} dx = -5$$

$$\Rightarrow \int_4^\infty f'(x)e^{\frac{-x}{4}} dx = -\frac{5}{4}-\frac{14}{e}$$


