can someone help me find the antiderivative of e^(4x^(2))?

Question

anonymous · Answer

what do you think

anonymous · Answer

What is the whole problem?

anonymous · Answer

well i tried substitution but it got a little complicated with the integral. what i needed to do was find the integral of $$\pi \int\limits_{0}^{1}  e ^{4}- e ^{4x ^{2}}dx$$ and i know what to do with everything but finding the aniderivative of $$e ^{4x ^{2}}$$ was getting diffucult

anonymous · Answer

let:
$$u = x^2$$

$$\sqrt{u} = x$$

du = 2xdx

du/x = 2dx

$$du/\sqrt{u} = 2dx$$

so now:

$$\int\limits_{}^{} e ^{x^2} dx = (1/2)\int\limits_{}^{} (e^u)(1/\sqrt{u}) du$$

and then i got stuck :(

anonymous · Answer

why does u=$$x^{2}$$ and not 4$$x ^{2}$$?

amistre64 · Answer

e^(x^2) has no elementary function to find

anonymous · Answer

$$4x ^{2}$$

amistre64 · Answer

unless you got something to counteract that 4x^2 its not doable with what youve been taguth

amistre64 · Answer

taught .. lol, at least i got the right keys

anonymous · Answer

do you all want the very very original problem? and np lol

anonymous · Answer

please

amistre64 · Answer

if ther eis an original then yes

anonymous · Answer

okay hold on it's a little long. thank you all!

anonymous · Answer

welcome

anonymous · Answer

Sketch the region R bounded by the graphs of the given equations. After showing a typical cross-sectional, find the volume of the solid generated by revolving R about the x-axis.
$$y=e ^{2x}$$
x=0
$$y=e ^{2}$$

amistre64 · Answer

ugh, this one all over again

amistre64 · Answer

or something close to it lol

anonymous · Answer

and after that i did 
$$\pi \int\limits_{0}^{1} r _{1}^{2} - r _{2}^{2}dx$$

amistre64 · Answer

$$(e^{2x})^2=e^{2x}*e^{2x}=e^{2x+2x}=e^{4x }$$

anonymous · Answer

then pluged it in to get $$\pi \int\limits_{0}^{1} (e ^{2})^{2} - (e ^{2x})^{2}dx$$

anonymous · Answer

yeah and then i got stuck at finding the antiderivative of $$e ^{4x ^{2}}$$

amistre64 · Answer

then quit trying to find the antiderivative of a function you dont even need

anonymous · Answer

to find the integral im suppose to

amistre64 · Answer

look up like 5 posts

amistre64 · Answer

see the difference between what your doing and what your spose to do?

anonymous · Answer

so when i plugged it in it's suppose to be multiplied by 2 and not squared?

amistre64 · Answer

when you power up exponents; the result is they multiply together

amistre64 · Answer

$$(b^n)^p=b^{np}$$

anonymous · Answer

LOL oh man:

$$r = e ^{2x}$$
then:
$$r^2 = e ^{4x}$$

THAT'S A LOT EASIER TO INTEGRATE

anonymous · Answer

oh okay no wonder!!! >.< thank you thank you both!!

amistre64 · Answer

youre welcome :)