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Math Help - Integration

  1. #1
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    Integration

    Hi i am completely stuck trying to integrate the following function.
    f(x) = (x + 2)(1 - e^ -x/3).
    Do i try and expand the brackets before i start?
    I tried doing it as a substitution using 1-e^-x/3 as u and
    du = 1/3 e^ -1/3x but that isn't one constant away from the original function when i multiply.

    Any help would be brilliant.

    Thanks

    Nikki
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by NIKKI020 View Post
    Hi i am completely stuck trying to integrate the following function.
    f(x) = (x + 2)(1 - e^ -x/3).
    Do i try and expand the brackets before i start?
    I tried doing it as a substitution using 1-e^-x/3 as u and
    du = 1/3 e^ -1/3x but that isn't one constant away from the original function when i multiply.

    Any help would be brilliant.

    Thanks

    Nikki
    \int {\left( {x + 2} \right)\left( {1 - {e^{ - x/3}}} \right)} dx = \int {\left( {x + 2} \right)} d\left( {x + 3{e^{ - x/3}}} \right) =

    = \left( {x + 2} \right)\left( {x + 3{e^{ - x/3}}} \right) - \int {\left( {x + 3{e^{ - x/3}}} \right)d\left( {x + 2} \right)}  =

    = \left( {x + 2} \right)\left( {x + 3{e^{ - x/3}}} \right) - \int {\left( {x + 3{e^{ - x/3}}} \right)dx}  =

    = \left( {x + 2} \right)\left( {x + 3{e^{ - x/3}}} \right) - \frac{{{x^2}}}{2} - 3\int {{e^{ - x/3}}dx}  =

    = \left( {x + 2} \right)\left( {x + 3{e^{ - x/3}}} \right) - \frac{{{x^2}}}{2} + 9\int {d\left( {{e^{ - x/3}}} \right)}  =

    = \left( {x + 2} \right)\left( {x + 3{e^{ - x/3}}} \right) - \frac{{{x^2}}}{2} + 9{e^{ - x/3}} + C =

    = \frac{{{x^2}}}{2} + 2x + 3x{e^{ - x/3}} + 15{e^{ - x/3}} + C.
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  3. #3
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    Thanks for that, i was looking at it completely the wrong way. Cheers.

    Nikki
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by NIKKI020 View Post
    Thanks for that, i was looking at it completely the wrong way. Cheers.

    Nikki
    Do you understand the essence of the method of integration by parts?
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  5. #5
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    (Sorry DeMath, if this is far from what you had in mind! But anyway...)

    Just in case a picture helps...



    (As usual, straight lines differentiate downwards or integrate up.)

    The essence of integration by parts is to work backwards through the product rule for differentiation...



    ... we want to fill out this product-rule shape usefully, but starting in one of the bottom corners instead of - as happens in differentiation, the top one. Choosing the right arrangement can be tricky, but once you've started in the right way, as above, it's really a matter of following through. Fill in the rest of the product-rule shape, then you'll need to cancel the addition you just entered, to make the lower equals sign valid, again.

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  6. #6
    Senior Member DeMath's Avatar
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    Tom, I have not absolutely understood, to whom your post is addressed: for me or for NIKKI020.
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  7. #7
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    Sorry again DeMath - I was being cheeky and crashing your gig!* Hope you don't mind. Did I get to the 'essence', though?


    *Taking the stage during your appearance.
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  8. #8
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    Yeah i have done a bit of it but i think i was making the problem over complicated! went back to my books to recap on method after you sent the steps to make sure i understood it. Think i was confusing it with integration by substitution. Thanks for your help!

    Nikki
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  9. #9
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    Thanks for that, i haven't seen it done like that before. It makes it a bit simpler! Cheers
    Nikki

    Quote Originally Posted by tom@ballooncalculus View Post
    (Sorry DeMath, if this is far from what you had in mind! But anyway...)

    Just in case a picture helps...



    (As usual, straight lines differentiate downwards or integrate up.)

    The essence of integration by parts is to work backwards through the product rule for differentiation...



    ... we want to fill out this product-rule shape usefully, but starting in one of the bottom corners instead of - as happens in differentiation, the top one. Choosing the right arrangement can be tricky, but once you've started in the right way, as above, it's really a matter of following through. Fill in the rest of the product-rule shape, then you'll need to cancel the addition you just entered, to make the lower equals sign valid, again.

    _______________________________

    Don't integrate - balloontegrate!

    Balloon Calculus Forum
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  10. #10
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    " Don't integrate - balloontegrate! "
    Best tip I've ever seen and a really awesome website, thanks for that ;D
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