Improper integrals

hecaterc

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Joined
Apr 11, 2019
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Pls help me with this question attached below. Thx a lot.

image.jpg
 
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Hello, and welcome to FMH! :)

I would begin with:

[MATH]\int_4^{\infty} f(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
Using integration by parts, where:

[MATH]u=f(x)\implies du=f'(x)\,dx[/MATH]
[MATH]dv=e^{-\frac{x}{7}}\,dx\implies v=-7e^{-\frac{x}{7}}[/MATH]
We may then write:

[MATH]\left[-7f(x)e^{-\frac{x}{7}}\right]_4^{\infty}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
Can you proceed?
 
Hello, and welcome to FMH! :)

I would begin with:

[MATH]\int_4^{\infty} f(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
Using integration by parts, where:

[MATH]u=f(x)\implies du=f'(x)\,dx[/MATH]
[MATH]dv=e^{-\frac{x}{7}}\,dx\implies v=-7e^{-\frac{x}{7}}[/MATH]
We may then write:

[MATH]\left[-7f(x)e^{-\frac{x}{7}}\right]_4^{\infty}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
Can you proceed?

Yes. I have got the answer. Thx a lot!
 
So that others reading this thread can see the problem worked to completion, I observed that given:

[MATH]|f(x)|<x^9+7[/MATH]
We must have:

[MATH]\lim_{x\to\infty}f(x)e^{-\frac{x}{7}}=0[/MATH]
And so the equation becomes:

[MATH]-7\left(0-f(4)e^{-\frac{4}{7}}\right)+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
[MATH]7f(4)e^{-\frac{4}{7}}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
Given:

[MATH]f(4)=-11[/MATH]
We may write:

[MATH]-77e^{-\frac{4}{7}}+7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5[/MATH]
[MATH]7\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=-5+77e^{-\frac{4}{7}}[/MATH]
[MATH]\int_4^{\infty} f'(x)e^{-\frac{x}{7}}\,dx=11e^{-\frac{4}{7}}-\frac{5}{7}[/MATH]
 
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