- Thread starter hecaterc
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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!

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?

- Joined
- Nov 24, 2012

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

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:

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

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

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