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Thread: Integration question

  1. #1
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    Integration question

    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
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    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 $\displaystyle u = 2x^4$ will do

    can you proceed?
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  3. #3
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    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.
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  4. #4
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    Just let:

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

    Make the subs and it'll fall into place.
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  5. #5
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    i see what you mean kind of! but it isnt really working!!! please wouls someone help me go through all the steps.
    Thankyouuuu
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  6. #6
    TD!
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    $\displaystyle
    \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}
    $
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    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

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

    Let $\displaystyle u = 2x^4$

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

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

    So our integral becomes:

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

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


    TD!'s underbraces make the substitution more explicit
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  8. #8
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    Quote Originally Posted by curlywurlysqurly View Post
    integrate: 4x^3 exp(2x^4)
    Note that $\displaystyle (e^{2x^4})'=8x^3e^{2x^4}$

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

    Divide by two and yields

    $\displaystyle \frac12e^{2x^4}+k=\int4x^3e^{2x^4}\,dx$
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  9. #9
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    Thumbs up

    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!
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  10. #10
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    Sorry, that was actually to Jhevon but thanks to everyone!!
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