Integration question

curlywurlysqurly · #1 $Old$ October 21st, 2007, 02:42 PM

Hey, i was wondering if anyone could help me with this question:

integrate: 4x^3 exp(2x^4)

i know what the answer is but i really dont understand how to do the question as apparently, it doesnt involve any integration at all!!!

Thanks

Jhevon · #2 $Old$ October 21st, 2007, 02:45 PM

Quote:

Originally Posted by curlywurlysqurly $View Post$

Hey, i was wondering if anyone could help me with this question:

integrate: 4x^3 exp(2x^4)

i know what the answer is but i really dont understand how to do the question as apparently, it doesnt involve any integration at all!!!

Thanks

a simple substitution of $u = 2x^4$ will do

can you proceed?

curlywurlysqurly · #3 $Old$ October 21st, 2007, 02:48 PM

sorry, no! i still dont understand. I was told to differentiate the (2x^4) and then put the whole thing over the (8x^3), it then simplifies, and you dont do any integration.

galactus · #4 $Old$ October 21st, 2007, 02:49 PM

Just let:

$u=2x^{4}, \;\ du=8x^{3}, \;\ \frac{du}{8}=x^{3}$

Make the subs and it'll fall into place.

curlywurlysqurly · #5 $Old$ October 21st, 2007, 03:09 PM

i see what you mean kind of! but it isnt really working!!! please wouls someone help me go through all the steps.
Thankyouuuu

TD! · #6 $Old$ October 21st, 2007, 03:13 PM

Jhevon · #7 $Old$ October 21st, 2007, 03:14 PM

Quote:

Originally Posted by curlywurlysqurly $View Post$

i see what you mean kind of! but it isnt really working!!! please wouls someone help me go through all the steps.
Thankyouuuu

fine, we will spoil you this time

$\int 4x^3 e^{2x^4}$

Let $u = 2x^4$

$\Rightarrow du = 8x^3 ~dx$

$\Rightarrow \frac {du}2 = 4x^3 ~dx$

So our integral becomes:

$\frac 12 \int e^u~du = \frac 12 e^u + C$

$= \frac 12 e^{2x^4} + C$

TD!'s underbraces make the substitution more explicit

Krizalid · #8 $Old$ October 21st, 2007, 03:16 PM

Quote:

Originally Posted by curlywurlysqurly $View Post$

integrate: 4x^3 exp(2x^4)

Note that $(e^{2x^4})'=8x^3e^{2x^4}$

Integrate both sides $e^{2x^4}+k_1=\int8x^3e^{2x^4}\,dx$

Divide by two and yields

$\frac12e^{2x^4}+k=\int4x^3e^{2x^4}\,dx$

curlywurlysqurly · #9 $Old$ October 21st, 2007, 03:20 PM

Ooooohhhhh thankyou! very clever, now that you have done it i remember doin that at school! lol ahh well, needed that for my test tomorrow!!!
Thanks!

curlywurlysqurly · #10 $Old$ October 21st, 2007, 03:26 PM

Sorry, that was actually to Jhevon but thanks to everyone!!