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    • Thread Starter
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    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.
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    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.
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    Do you know what e^{(x^3)} differentiates to? If so, this question can be done easily by inspection.
    • Thread Starter
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    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
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    Id do what JohnnySPal tells you.

    and BTW

    if  y = e^{f(x)}

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

    so e^{x^3} doesnt differentiate to 3(x^2)*(e^(x^3))/8x^3)
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    (Original post by Dave_McDougall)
    I believe e^(x^3) differentiates to 3(x^2)*(e^(x^3))/8x^3)
    No, \frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)}

    So

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

    /edit: humbug, beaten to it. >_>
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    *sexes insparato*
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    so I did what Johnny said and got as an answer:
    \frac{1}{3{x^2}} e^{x^3}
    right?
    edit: if I could actually differentiate I could check >.<
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    Err... No...

    After the substitution you should get the integral

    \int \frac{1}{3} e^u du

    So after intagration and converting back to x you get:

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

    Unless I'm being thick?
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    (Original post by JohnnySPal)
    Err... No...

    After the substitution you should get the integral

    \int \frac{1}{3} e^u du

    So after intagration and converting back to x you get:

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

    Unless I'm being thick?
    Nope looks right to me.
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    (Original post by Dave_McDougall)
    so I did what Johnny said and got as an answer:
    \frac{1}{3{x^2}} e^{x^3}
    right?
    edit: if I could actually differentiate I could check >.<
    \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.
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    (Original post by JohnnySPal)
    Err... No...

    After the substitution you should get the integral

    \int \frac{1}{3} e^u du

    So after intagration and converting back to x you get:

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

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