Integration question

**curlywurlysqurly** · October 21st 2007, 12:42 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** · October 21st 2007, 12:45 PM

Originally Posted by curlywurlysqurly

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** · October 21st 2007, 12:48 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** · October 21st 2007, 12:49 PM

Just let:

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

Make the subs and it'll fall into place.

**curlywurlysqurly** · October 21st 2007, 01:09 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!** · October 21st 2007, 01:13 PM

$<br /> \int {4x^3 e^{2x^4 } \,\mbox{d}x} = \frac{1}{2}\int {\underbrace {e^{2x^4 } }_{e^u }\underbrace {8x^3 \,\mbox{d}x}_{\rm{d}u}} = \frac{1}{2}\int {e^u \,\mbox{d}u} <br />$

**Jhevon** · October 21st 2007, 01:14 PM

Originally Posted by curlywurlysqurly

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** · October 21st 2007, 01:16 PM

Originally Posted by curlywurlysqurly

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** · October 21st 2007, 01:20 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** · October 21st 2007, 01:26 PM

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

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