Should I have multiplied by -2 below?

$\displaystyle \int\displaystyle\frac{y}{e^-2y}dy$

$\displaystyle =\displaystyle\frac{-2.y^2.ln e^-2y}{2}+C$

$\displaystyle =-y^2.ln e^-2y+C$

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- Nov 8th 2006, 03:28 PMashuraIntegrate 1/e^-2y
Should I have multiplied by -2 below?

$\displaystyle \int\displaystyle\frac{y}{e^-2y}dy$

$\displaystyle =\displaystyle\frac{-2.y^2.ln e^-2y}{2}+C$

$\displaystyle =-y^2.ln e^-2y+C$ - Nov 8th 2006, 03:43 PMtopsquark
I'm confused about something here. Why aren't you writing this as:

$\displaystyle \int \, ye^{2y} dy$

That, at least, would get rid of the pesky negative sign. Also, I don't know where you are getting the "ln" from. This can be done using integration by parts:

$\displaystyle \int u dv = uv - \int v du$

So let $\displaystyle u = y$ and $\displaystyle dv = e^{2y}dy$

Then

$\displaystyle du = dy$ and $\displaystyle v = \frac{1}{2}e^{2y}$

So

$\displaystyle \int \, ye^{2y} dy = \frac{1}{2}ye^{2y} - \int \frac{1}{2}e^{2y}dy$

and

$\displaystyle \int \frac{1}{2}e^{2y}dy = \frac{1}{4}e^{2y}$

Thus

$\displaystyle \int \, ye^{2y} dy = \frac{1}{2}ye^{2y} - \frac{1}{4}e^{2y} + C$ (Don't forget the integration constant!)

-Dan - Nov 8th 2006, 03:46 PMtopsquark
A couple of tips about the LaTeX coding:

First, I don't know what the \displaystyle tag is supposed to do, but it doesn't seem to be doing anything on this Forum.

Also, if you want to write an exponent with more than one character, enclose the exponent in { }. Example:

$\displaystyle e^2y$ (No { }.)

$\displaystyle e^{2y}$ (With the { }.)

Hint: You can see how others have coded their LaTeX by quoting the post. You will see the coding in the quote box.

-Dan - Nov 8th 2006, 04:40 PMashura
I used the form,

$\displaystyle \int\frac{y}{e^{-2y}}dy $

because question was such. Also

$\displaystyle \int\frac{1}{e^{-2y}} = $

$\displaystyle -2ln e^{-2y} + C$.

Agree that

$\displaystyle \int y.e^{2y}dy$

is easier to work with. Thanks for the formatting info. - Nov 9th 2006, 03:38 AMtopsquark
- Nov 9th 2006, 09:42 AMashura
$\displaystyle \int \frac{1}{e^{-2y}}dy = \int e^{2y}dy$,

let $\displaystyle u = 2y$

$\displaystyle \frac{du}{dy} = 2$

$\displaystyle dy = \frac{du}{2}$

substituting for u,

$\displaystyle \int e^{2y}dy = \int e^u \frac{du}{2}$

$\displaystyle = \frac{1}{2}\int e^u{du}$

$\displaystyle = \frac{1}{2}e^u+C$

substituting for u,

$\displaystyle = \frac{1}{2}e^{2y}+C$

Thanks.