integral with e to the sq. x

**chocolatelover** · September 4th 2007, 09:05 AM

This is what I did:

u=sq. x
2du=1/2sq.x dx

2 int. 1 to 4 ue to the u du

then integration by parts:

y=u
dv=e to the u dy
du=u2/2
v=e to the u

=ue to the u-2 int. 1 to 4 e to the uu2/2
=ue to the u-2|u2eu-eu2u (1 to 4)
=ue to the u-16e4-6e (after I pluged in 4 and 1)

Can someone please show me what I did wrong?

Thank you

**topsquark** · September 4th 2007, 11:10 AM

Originally Posted by chocolatelover

This is what I did:

$u = \sqrt{x}$

$2du=\frac{1}{2\sqrt{x}} dx$

$2 \int_1^4 ue^u du$

then integration by parts:

$y=u$
$dv=e^u dy$
$du=\frac{u2}{2}$
$v=e^u$

I have retyped some of your previous post in LaTeX.

First, where did the 2 in the $2du$ (second line) come from?

Also your notation for the integration by parts is confusing. And what is this u2 thing?

-Dan

**topsquark** · September 4th 2007, 11:17 AM

$\int e^{\sqrt{x}} ~dx$

Let's change the variable to something like y instead of u (which is used in the integration by parts formula.)

Let
$y = \sqrt{x} \implies dy = \frac{1}{2\sqrt{x}}dx$

So
$dy = \frac{1}{2y}dx$
or
$dx = 2y dy$

So
$\int e^{\sqrt{x}} ~dx = \int e^y \cdot 2y dy = 2\int ye^y dy$

Now let's do the integration by parts.
Let
$u = y \implies du = dy$
and
$dv = e^y ~ dy \implies v = e^y$

So
$\int ye^y ~ dy = ye^y - \int e^y ~ dy$

Now you take it from here. And remember, bookkeeping is important! Keep track of the details; this is where you seem to have most of your problems.

-Dan

**chocolatelover** · September 4th 2007, 05:39 PM

The integral is from 1 to four. After what you gave me I put:

=ye to the y-2|1 to 4 e to the y
=ye to the y-2(e to the 4tj -e to the 1st)
=ye to the y-2e to the 4th+2e

Now, do I plug in y with sq. x?

Thank you

**Jhevon** · September 4th 2007, 07:54 PM

Originally Posted by Krizalid

Jhevon, it's a definite one

of course it is, i was just seeing if chocolatelover was paying attention. i leave plugging in the limits to him/her

**chocolatelover** · September 4th 2007, 08:10 PM

The integral is from 1 to four. After what you gave me I put:

=ye to the y-2|1 to 4 e to the y
=ye to the y-2(e to the 4tj -e to the 1st)
=ye to the y-2e to the 4th+2e

Now, do I plug in y with sq. x?

So, my final answer would be sq.xe to sq.x-2e4+2e+c???

Thank you

**Jhevon** · September 4th 2007, 08:13 PM

Originally Posted by chocolatelover

The integral is from 1 to four. After what you gave me I put:

=ye to the y-2|1 to 4 e to the y
=ye to the y-2(e to the 4tj -e to the 1st)
=ye to the y-2e to the 4th+2e

Now, do I plug in y with sq. x?

So, my final answer would be sq.xe to sq.x-2e4+2e+c???

Thank you

definite integral means you should have a numerical answer. if you have an x in the final answer, you're wrong

look on the previous page for the answer i gave. plug the limits into that

**topsquark** · September 5th 2007, 06:35 AM

Originally Posted by chocolatelover

The integral is from 1 to four. After what you gave me I put:

=ye to the y-2|1 to 4 e to the y
=ye to the y-2(e to the 4tj -e to the 1st)
=ye to the y-2e to the 4th+2e

Now, do I plug in y with sq. x?

So, my final answer would be sq.xe to sq.x-2e4+2e+c???

Thank you

If you are going to be writing a lot of these (and it seems that you are) it might be fruitful for you to look into how to write these using LaTeX. See the beginning of this thread. You can also left click on the equations and a box will come up telling you what the code is.

-Dan

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