# 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: $$\int_4^\infty f\,'(x)e^{-x/4} dx\;.$$

• Try integration by parts and use the fact that the given integral value to be $-5$ and $f(4)=14 ,|f(x)| < x^3+10$, then you should get $\frac{14}{e}$ as your answer Commented Mar 18, 2012 at 16:53
• @Kirthi: I think it should be $-\frac{14}{e}-\frac{5}{4}$.
– TMM
Commented Mar 18, 2012 at 17:10
• I think it is rather $\frac{14}{e} + \frac{5}{4}$ Commented Mar 18, 2012 at 17:11
• You need to assume more than just continuity here. To use integration by parts (or the Fundamental Theorem of Calculus), you want $f(x)$ to be absolutely continuous. Commented Mar 18, 2012 at 17:33

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}

Hint: Use integration by parts.

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 $$\frac{4f(14)}{e} + 4 \int f'(x)e^{\frac{-x}{4}} dx = -5$$ $$\Rightarrow \int_4^\infty f'(x)e^{\frac{-x}{4}} dx = -\frac{5}{4}-\frac{14}{e}$$
• Missed dx in few places, so edited that Commented Mar 18, 2012 at 19:12