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    Hello again! Could someone suggest what method I should use to solve this question:

    1) integral (0.1 -> 0) x^2.e^x^3 dx

    I tried using by parts twice but I got stuck the second time round as it didnt seem to have made the integral any easier. (i am right in thinking that integral of e^x^3 = [1/3x^2]e^x^3?)

    Secondly, could someone work through the below problem as I end up with something completely different to the answers! arghh:

    2) Find, in terms of e and pi, the volume generatre when the region bounded by the lines x=1, x=3 and y=0 and the curve with equation y=e^[3x/2] is rotated completely about the x-axis.

    Thank you all so much - i figured you wouldn't mind too much as you'll enjoy the P3 practise ready for next week!
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    (Original post by Hoofbeat)
    2) Find, in terms of e and pi, the volume generatre when the region bounded by the lines x=1, x=3 and y=0 and the curve with equation y=e^[3x/2] is rotated completely about the x-axis.

    Thank you all so much - i figured you wouldn't mind too much as you'll enjoy the P3 practise ready for next week!
    For 2), I get V = (1/3)(pi)e³[ e³ - 1] ???
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    1. Spot that the derivative of x^3 = 3x^2 so use the substitution u=x^3, dx = du/(3x^2)

    I don't think it's possible to integrate e^x^3 - when you differentiate [1/3x^2]e^x^3 you'd need to use the quotient rule as well as the chain rule.
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    (Original post by mockel)
    For 2), I get V = (1/3)(pi)e³[ e³ - 1] ???
    I get that too
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    (Original post by Hoofbeat)
    Hello again! Could someone suggest what method I should use to solve this question:

    1) integral (0.1 -> 0) x^2.e^x^3 dx
    I tried using the substitution u=x³ , and it seems to help (cancels the x^2) term...so maybe that would be an appropriate way to go...

    Edit: just saw your posts bezza.....glad to know that I seem to be hitting my peak just before the exam
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    well integrating e^x^3 is ok since u can let u = x^3 and du/dx = 3x^2 therefore reaaraging gives 1/3du = x^2 dx.

    so this means integrating 1/3e^u which is 1/3e^u (u = x^3) so this becomes 1/3e^x^3. sticking the limits in gives 1/3e^1/1000 - 1/3. put 1/3 outside into a bracket and becomes 1/3(e^0.001 - 1).

    ?? plizzz say dats ryt!
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    (Original post by Jubba)
    well integrating e^x^3 is ok since u can let u = x^3 and du/dx = 3x^2 therefore reaaraging gives 1/3du = x^2 dx.

    so this means integrating 1/3e^u which is 1/3e^u (u = x^3) so this becomes 1/3e^x^3. sticking the limits in gives 1/3e^1/1000 - 1/3. put 1/3 outside into a bracket and becomes 1/3(e^0.001 - 1).

    ?? plizzz say dats ryt!
    I think that should be right
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    (Original post by Hoofbeat)
    Hello again! Could someone suggest what method I should use to solve this question:

    1) integral (0.1 -> 0) x^2.e^x^3 dx

    I tried using by parts twice but I got stuck the second time round as it didnt seem to have made the integral any easier. (i am right in thinking that integral of e^x^3 = [1/3x^2]e^x^3?)

    Secondly, could someone work through the below problem as I end up with something completely different to the answers! arghh:

    2) Find, in terms of e and pi, the volume generatre when the region bounded by the lines x=1, x=3 and y=0 and the curve with equation y=e^[3x/2] is rotated completely about the x-axis.

    Thank you all so much - i figured you wouldn't mind too much as you'll enjoy the P3 practise ready for next week!
    the first integrates perfectly to 1/3 e^x^3. differentiate it and see for yourself.

    the second one is basically integrate between 1 and 3, pi y^2 dx.

    so it's the integral of pi e^3x = 1/3 pi e^3x, between limits 1 and 3. so your answer should be 1/3 pi (e^9 - e^3).

    i hope that's right
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    could someone work through the below problem
    I can indeed.

    The volume of a rotated shape is the (pi) x ∫y² dx

    This gives (pi) x ∫(e^(3x/2))^2 dx (Between limits 1 and 3)

    = (pi) x ∫e^(3x) dx

    = (pi)[(1/3)(e^3x)] between 1 and 3

    Take out the factors 1/3 and e^3 gives::

    (1/3)(pi)e³[ e^6 - 1]

    I think bezza and mockel have accidently put e^3 in the brackets, but e^9 = e^3 x e^6, and not e^3 x e^3!

    Hope this helps
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    (Original post by 4Ed)
    the first integrates perfectly to 1/3 e^x^3. differentiate it and see for yourself.

    the second one is basically integrate between 1 and 3, pi y^2 dx.

    so it's the integral of pi e^3x = 1/3 pi e^3x, between limits 1 and 3. so your answer should be 1/3 pi (e^9 - e^3).

    i hope that's right
    yeah...sorry, integrated the first one from 1 to 2 for some reason :confused:
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    (Original post by SUKBarracuda)
    I think bezza and mockel have accidently put e^3 in the brackets, but e^9 = e^3 x e^6, and not e^3 x e^3!
    Oops! yeah, that's what I meant - I was just copying mockel - it's all his fault!
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    (Original post by SUKBarracuda)
    I can indeed.

    The volume of a rotated shape is the (pi) x ∫y² dx

    This gives (pi) x ∫(e^(3x/2))^2 dx (Between limits 1 and 3)

    = (pi) x ∫e^(3x) dx


    = (pi)[(1/3)(e^3x)] between 1 and 3

    Take out the factors 1/3 and e^3 gives::

    (1/3)(pi)e³[ e^6 - 1]

    I think bezza and mockel have accidently put e^3 in the brackets, but e^9 = e^3 x e^6, and not e^3 x e^3!

    Hope this helps
    Thanks for all your help. However, i am slightly confused. surely 3^[3x/2] all squared would give e^[9x^2/4]? Oh god, how am I going to get my A in P3 if i can't even do simple indices!!!
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    (Original post by Hoofbeat)
    Thanks for all your help. However, i am slightly confused. surely 3^[3x/2] all squared would give e^[9x^2/4]? Oh god, how am I going to get my A in P3 if i can't even do simple indices!!!
    One of the rules of indices is (x^y)^z = x^(yz). I assume you meant [e^(3x/2)]^2 which equals e^(3x/2*2) = e^3x
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    Bezza is right. e^(3x/2) all squared is the same as:

    e^(3x/2) x e^(3x/2)

    which is e^((3x/2) x 2)

    which is e^(3x).

    You can think of the indice theory in terms of 2s.

    2^4 = 16

    so (2^4) all squared is 16 x 16 which equals 256.

    and this is equal to 2^8, which is of course the same as 2^(4x2).

    But don't worry you'll do fine in the exam if you just practice a bit of algebra over the next week.
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    (Original post by SUKBarracuda)
    Bezza is right. e^(3x/2) all squared is the same as:

    e^(3x/2) x e^(3x/2)

    which is e^((3x/2) x 2)

    which is e^(3x).

    You can think of the indice theory in terms of 2s.

    2^4 = 16

    so (2^4) all squared is 16 x 16 which equals 256.

    and this is equal to 2^8, which is of course the same as 2^(4x2).

    But don't worry you'll do fine in the exam if you just practice a bit of algebra over the next week.
    Thanks both of you! lol
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    This is related to integration (i promise!!! !!!)

    Ok in the above question, you have to integrate e^(3x), so i was wondering what is the integral of something like e^(3(x^2))

    Thanks
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    (Original post by Silly Sally)
    This is related to integration (i promise!!! !!!)

    Ok in the above question, you have to integrate e^(3x), so i was wondering what is the integral of something like e^(3(x^2))

    Thanks
    u cannot express the integral of e^x^2 in terms of elementary functions,but u can find the exact value of the area under the graph of y=e^-x^2.
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    (Original post by IntegralAnomaly)
    u cannot express the integral of e^x^2 in terms of elementary functions,but u can find the exact value of the area under the graph of y=e^-x^2.
    Sorry, don't understand what you mean.
 
 
 

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