# Thread: integral with e to the sq. x

1. ## integral with e to the sq. x

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

2. 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 $\displaystyle \int e^{\sqrt {x}}~dx = \int \frac {2 \sqrt {x}}{2 \sqrt {x}}e^{\sqrt {x}}~dx$

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

3. 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

4. $\displaystyle \int e^{\sqrt x}\,dx$

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

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

Next step, it's integration by parts.

5. 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

6. 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

7. I suggest you get practice on LaTeX, not hard.

8. 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 $\displaystyle 2 \int u e^u~du$. where did $\displaystyle \sqrt {x}$ come from? pay attention

9. 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?

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

11. Originally Posted by Krizalid
Better you set $\displaystyle y=u$ & $\displaystyle dv=e^u\,du$
yeah, the u's will get mixed up. use t for the first substitution.

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

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

12. 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

13. Is the final answer ueu-16e4-6e

14. Originally Posted by chocolatelover
the final answer is $\displaystyle 2 e^{\sqrt {x}}(\sqrt {x} - 1)$