# Integral

• May 14th 2008, 11:03 AM
Integral
Hello

How do you integrate this, Integral (5x/3)*e^-x/3

Thank you
• May 14th 2008, 11:07 AM
Moo
Hello,

Quote:

Hello

How do you integrate this, Integral (5x/3)*e^-x/3

Thank you

When you have the integral of a product involving a polynomial (5x/3) and a "basic" exponential (e^-x/3), think "Integration by parts".

Take $u'(x)=\text{exponential thing}$ and $v(x)=\text{polynomial thing}$
• May 14th 2008, 11:11 AM
You refer to partiall integration?

F(x)g(x) - integral F(x)g(x)´

Så I should set F(x) = 5x/3 and g(x) = e^-x/3 ?

thank you
• May 14th 2008, 11:14 AM
Moo
Quote:

You refer to partiall integration?

F(x)g(x) - integral F(x)g(x)´

Så I should set F(x) = 5x/3 and g(x) = e^-x/3 ?

thank you

Hm...

This is not the formula, but it looks like...

$\int_a^b f(x)g'(x) dx=\left[f(x)g(x)\right]_a^b - \int_a^b f'(x)g(x) dx$

And take $f(x)=5x/3$ and $g'(x)=e^{-x/3}$

(Sun)

Edit : your writing is confusing me... What is the original integral with f and g for you ?
• May 14th 2008, 11:26 AM
The original integral is "integral 5x/3*e^-x/3"

I set g(x) = e^-x/3
And f(x) = 5x/3

then used the formula

F(x)g(x) - integral F(x)g(x)´
• May 14th 2008, 11:33 AM
Moo
Quote:

The original integral is "integral 5x/3*e^-x/3"

I set g(x) = e^-x/3
And f(x) = 5x/3

then used the formula

F(x)g(x) - integral F(x)g(x)´

Ok that's it :)

Set f(x)=e^-x/3
And g(x)=5x/3

Because g will be differentiated. Since the derivative of an exponential is in most case the same, it's not really useful to set g(x)=e^-x/3
Whereas if you differentiate a polynomial (5x/3), you will soon get to a constant.
Do you understand ?

(Wink)
• May 14th 2008, 11:37 AM