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integration by substitution problem - what do I substitute?

Hey,
so I am sitting here doing homework, and I am stuck. As in seriously stuck. I have tried this question for an hour, in a million different ways, and it doesn't work for me. Maybe it just doesn't like me or something:p:
anyway, I need your help with this. Here goes:
use integration by substitution to work out:
∫(x^2)e^(x^3)dx

It's fine with the x^2 part and e^x, but the e^(x^3) just completely kills me and I have no clue as to how to integrate the whole expression.

I'd greatly appreciate any pointers :yep:

Thanks, guys.
Reply 1
Avatar for JohnnySPal
JohnnySPal
2
Try u = x^3. When you work out "du" you'll find that the x^2 disappears nicely. Seriously, just work through with that substitution and you'll see it clears up nicely.
Reply 2
Avatar for BJack
BJack
19
Do you know what e(x3)e^{(x^3)} differentiates to? If so, this question can be done easily by inspection.
I believe e^(x^3) differentiates to 3(x^2)*(e^(x^3))/8x^3)

also, how do you use latex, this whole bracket and thingy stuff is really annoying and confusing.

Johnny, will try this now. thanks. will report in a minute
Reply 4
Avatar for insparato
insparato
15
Id do what JohnnySPal tells you.

and BTW

if y=ef(x) y = e^{f(x)}

dydx=f′(x)ef(x) \frac{dy}{dx} = f'(x)e^{f(x)}

so e^{x^3} doesnt differentiate to 3(x^2)*(e^(x^3))/8x^3)
Reply 5
Avatar for BJack
BJack
19
Dave_McDougall
I believe e^(x^3) differentiates to 3(x^2)*(e^(x^3))/8x^3)


No, ddxef(x)=f′(x)ef(x)\frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)}

So

ddxe(x3)=3x2e(x3)\frac{d}{dx} e^{(x^3)} = 3x^2e^{(x^3)}

/edit: humbug, beaten to it. >_>
Reply 6
Avatar for D-Day
D-Day
10
*sexes insparato*
so I did what Johnny said and got as an answer:
13x2ex3\frac{1}{3{x^2}} e^{x^3}
right?
edit: if I could actually differentiate I could check >.<
Reply 8
Avatar for JohnnySPal
JohnnySPal
2
Err... No...

After the substitution you should get the integral

∫13eudu\int \frac{1}{3} e^u du

So after intagration and converting back to x you get:

13ex3+c\frac{1}{3} e^{x^3} + c

Unless I'm being thick?
Reply 9
Avatar for Kyalimers
Kyalimers
21
JohnnySPal
Err... No...

After the substitution you should get the integral

∫13eudu\int \frac{1}{3} e^u du

So after intagration and converting back to x you get:

13ex3+c\frac{1}{3} e^{x^3} + c

Unless I'm being thick?

Nope looks right to me.
Dave_McDougall
so I did what Johnny said and got as an answer:
13x2ex3\frac{1}{3{x^2}} e^{x^3}
right?
edit: if I could actually differentiate I could check >.<

ddx13x2ex3=−6x−3ex3+ex3\dfrac{d}{dx} \dfrac{1}{3{x^2}} e^{x^3} = -6x^{-3}e^{x^3} + e^{x^3}

Immediately here you can tell that can't end up to what you'd want becausew you'd have to use the product rule if I recall correctly.
JohnnySPal
Err... No...

After the substitution you should get the integral

∫13eudu\int \frac{1}{3} e^u du

So after intagration and converting back to x you get:

13ex3+c\frac{1}{3} e^{x^3} + c

Unless I'm being thick?

ah, I know what I did wrong!
this I got: ∫13eudu\int \frac{1}{3} e^u du,
and then I made the mistake of saying 13uex3+c\frac{1}{3u} e^{x^3} + c,
i.e. thought that
ex+c e^{x} + c
would integrate to
exu\frac{e^x}{u}
which of course was wrong.
edit: I suck at latex
thanks everyone, I can now go to bed a happy chappy! Man, and this is only A2 normal!! how should I survive the FM?!?! :biggrin:

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