integral with e to the sq. x

**chocolatelover** · September 3rd 2007, 04:24 PM

Hi,

To solve the int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?

u=e to the x to the 1/2

du=e to the x(1-2x to the -1/2)

Is this correct?

Thank you

**Jhevon** · September 3rd 2007, 04:35 PM

Originally Posted by chocolatelover

Hi,

To solve the int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?

u=e to the x to the 1/2

du=e to the x(1-2x to the -1/2)

Is this correct?

Thank you

note that $\int e^{\sqrt {x}}~dx = \int \frac {2 \sqrt {x}}{2 \sqrt {x}}e^{\sqrt {x}}~dx$

now use the substitution, $u = \sqrt {x}$

**chocolatelover** · September 3rd 2007, 04:38 PM

How did you know that? Is it a rule?

Would du be e to sq. x(1/2x to the -1/2)dx??????????

Thank you

**Krizalid** · September 3rd 2007, 04:47 PM

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

$u=\sqrt x\implies du=\frac1{2\sqrt x}\,dx$ , the integral becomes to

$\int e^{\sqrt x}\,dx=2\int ue^u\,du$

Next step, it's integration by parts.

**Jhevon** · September 3rd 2007, 04:48 PM

Originally Posted by chocolatelover

How did you know that? Is it a rule?

Would du be e to sq. x(1/2x to the -1/2)dx??????????

Thank you

an explanation is given by TPH in the second post here

EDIT: Ah, I see Krizalid already answered

**chocolatelover** · September 3rd 2007, 04:57 PM

Would u=e to sq. x and dv=tdt????
du=e to the sq. x(1/2x to -1/2)
v=t2/2

Thank you

**Krizalid** · September 3rd 2007, 05:00 PM

I suggest you get practice on LaTeX, not hard.

This way, it'll be better to check your answers.

**Jhevon** · September 3rd 2007, 05:00 PM

Originally Posted by chocolatelover

Would u=e to sq. x and dv=tdt????
du=e to the sq. x(1/2x to -1/2)
v=t2/2

Thank you

what are you talking about? the integral you are doing by parts is $2 \int u e^u~du$ . where did $\sqrt {x}$ come from? pay attention

**chocolatelover** · September 3rd 2007, 05:27 PM

I got

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

then (e to the u)u2/2-2int.1 to 4(u2/2)e to the u

Now I can just plug in 1 and 4 right?

**Krizalid** · September 3rd 2007, 05:29 PM

Better you set $y=u$ & $dv=e^u\,du$

**Jhevon** · September 3rd 2007, 05:34 PM

Originally Posted by Krizalid

Better you set $y=u$ & $dv=e^u\,du$

yeah, the u's will get mixed up. use t for the first substitution.

so you want to find $2 \int t e^t~dt$

now go ahead

EDIT: and if you won't use LaTex, please write in math shorthand as i showed you in one of your earlier posts. reading through your work is annoying when you use sentences to describe math symbols

**chocolatelover** · September 3rd 2007, 05:38 PM

I got ue to the u-2int. 1 to 4 e to the u u2/2

Then I would do it again, right?

Thank you

**chocolatelover** · September 3rd 2007, 07:10 PM

Is the final answer ueu-16e4-6e

**Jhevon** · September 3rd 2007, 07:24 PM

Originally Posted by chocolatelover

Is the final answer ueu-16e4-6e

the final answer is $2 e^{\sqrt {x}}(\sqrt {x} - 1)$

**Krizalid** · September 4th 2007, 08:14 AM

Originally Posted by chocolatelover

int. 1 to 4 e to the sq. x dx you would first do a u substitution, right?

Jhevon, it's a definite one

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