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    Attachment 612190612192612200 any help with this question is appreciated... I think the bit but highlighted is what went wrong. Not sure how to solve this...

    Thanks
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    Name:  1484653180477-794039220.jpg
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    (Original post by coconut64)
    better picture .. thanks
    From \displaystyle \frac{2}{3} \int_{0}^{5} ue^u.du you should've done IBP with g=u and h'=e^u so I'm not sure what happened on the line right after that, it seems you messed up on the IBP.

    P.S. Factor the constants outside the integral, this makes life so much easier.
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    (Original post by RDKGames)
    From \displaystyle \frac{2}{3} \int_{0}^{5} ue^u.du you should've done IBP with g=u and h'=e^u so I'm not sure what happened on the line right after that, it seems you messed up on the IBP.

    P.S. Factor the constants outside the integral, this makes life so much easier.
    Is IBP the guess method ? Thanks
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    (Original post by coconut64)
    Is IBP the guess method ? Thanks
    Integration by parts.
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    (Original post by RDKGames)
    Integration by parts.
    I did try to do that but the power of u will keep increasing which doesn't really get you anywhere . How does that work ?
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    (Original post by coconut64)
    I did try to do that but the power of u will keep increasing which doesn't really get you anywhere . How does that work ?
    Yeah that's because you chose g=e^u, h'=u for IBP whereas you're supposed to choose g=u, h'=e^u which is doable and the integral collapses nicely.
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    (Original post by coconut64)
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    You got a lot of this question right - so should feel encouraged!

    Some tips

    When you choose u,if a squareroot is involved you have the option of letting (as here) u^2 = 3x+1 When you integrate, 2u du = 3 dx and so dx = 2/3 du

    Taking the constant outside the integral is definitely good advice!

    There's a simple rule of thumb worth remembering for integration by parts. It never pays to let u equal e to the power of x because you get stuck in a loop. It always pays to let ln x (log x) equal u because integrating ln x usually then makes progress.
 
 
 
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