Integrate 1/e^-2y

**ashura** · November 8th 2006, 03:28 PM

Should I have multiplied by -2 below?

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

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

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

**topsquark** · November 8th 2006, 03:43 PM

Originally Posted by ashura

Should I have multiplied by -2 below?

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

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

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

I'm confused about something here. Why aren't you writing this as:
$\int \, ye^{2y} dy$

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

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

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

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

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

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

-Dan

**topsquark** · November 8th 2006, 03:46 PM

Originally Posted by ashura

Should I have multiplied by -2 below?

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

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

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

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:
$e^2y$ (No { }.)

$e^{2y}$ (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

**ashura** · November 8th 2006, 04:40 PM

I used the form,

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

because question was such. Also

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

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

Agree that

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

is easier to work with. Thanks for the formatting info.

**topsquark** · November 9th 2006, 03:38 AM

Originally Posted by ashura

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

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

For the record:

$-2ln e^{-2y} + C = -2(-2y)ln(e) + C = 4y + C$

$\int\frac{1}{e^{-2y}}dy = \int e^{2y}dy = \frac{1}{2}e^{2y} + C$

As the two of these are not different by a constant, there is something wrong with how you integrated.

-Dan

**ashura** · November 9th 2006, 09:42 AM

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

let $u = 2y$

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

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

substituting for u,

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

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

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

substituting for u,

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

Thanks.

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