# Improper integrals

#### hecaterc

##### New member
Pls help me with this question attached below. Thx a lot.

Last edited by a moderator:

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

I would begin with:

$$\displaystyle \int_4^{\infty} f(x)e^{-\frac{x}{7}}\,dx=-5$$

Using integration by parts, where:

$$\displaystyle u=f(x)\implies du=f'(x)\,dx$$

$$\displaystyle dv=e^{-\frac{x}{7}}\,dx\implies v=-7e^{-\frac{x}{7}}$$

We may then write:

$$\displaystyle \left[-7f(x)e^{-\frac{x}{7}}\right]_4^{\infty}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5$$

Can you proceed?

#### hecaterc

##### New member
Hello, and welcome to FMH!

I would begin with:

$$\displaystyle \int_4^{\infty} f(x)e^{-\frac{x}{7}}\,dx=-5$$

Using integration by parts, where:

$$\displaystyle u=f(x)\implies du=f'(x)\,dx$$

$$\displaystyle dv=e^{-\frac{x}{7}}\,dx\implies v=-7e^{-\frac{x}{7}}$$

We may then write:

$$\displaystyle \left[-7f(x)e^{-\frac{x}{7}}\right]_4^{\infty}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5$$

Can you proceed?
Yes. I have got the answer. Thx a lot!

#### MarkFL

##### Super Moderator
Staff member
So that others reading this thread can see the problem worked to completion, I observed that given:

$$\displaystyle |f(x)|<x^9+7$$

We must have:

$$\displaystyle \lim_{x\to\infty}f(x)e^{-\frac{x}{7}}=0$$

And so the equation becomes:

$$\displaystyle -7\left(0-f(4)e^{-\frac{4}{7}}\right)+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5$$

$$\displaystyle 7f(4)e^{-\frac{4}{7}}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5$$

Given:

$$\displaystyle f(4)=-11$$

We may write:

$$\displaystyle -77e^{-\frac{4}{7}}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5$$

$$\displaystyle 7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5+77e^{-\frac{4}{7}}$$

$$\displaystyle \int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=11e^{-\frac{4}{7}}-\frac{5}{7}$$