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Math Help - Integration (Sort of)

  1. #1
    Member Awsom Guy's Avatar
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    Integration (Sort of)

    Find F(x) given f(x) is:
    (x+2)(x^2`+ 4x + 3)^5
    Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
    My attempt:
    u=(x^2 + 4x + 3)
    u'=2x+4
    u^5
    u^6/6 + C
    (x^2 + 4x + 3)^6/6 + C
    I should have 12 instead of 6 on the bottom.
    Answer should be:
    (x^2 + 4x + 3)^6/12 + C
    Thank you
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    Find F(x) given f(x) is:
    (x+2)(x^2`+ 4x + 3)^5
    Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
    My attempt:
    u=(x^2 + 4x + 3)
    u'=2x+4
    u^5
    u^6/6 + C
    (x^2 + 4x + 3)^6/6 + C
    I should have 12 instead of 6 on the bottom.
    Answer should be:
    (x^2 + 4x + 3)^6/12 + C
    Thank you
    You're forgetting the chain rule. Remember when you take take the derivative of that, you need to multiply it by the derivative of the inside. So, compensate for that when integrating by multiplying it by 1/(derivative of inside). The key here is noticing that the derivative of the inside (2x+4) is 2(x+2), notice the x+2 on the outside and you're all set. Feel free to ask me for any more help?
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  3. #3
    Member Awsom Guy's Avatar
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    Is that for integration.
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  4. #4
    Member mathemagister's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    Is that for integration.
    Since integration is just the opposite of differentiation, you need to do the opposite of the chain rule: instead of multiplying by the derivative of the inside, you need to divide by it. You can always check your answer by differentiating your answer and see if you get the question.
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  5. #5
    Member Awsom Guy's Avatar
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    I know that but how do I do it properly wait I'll try and then give you a buzz.
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  6. #6
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    Quote Originally Posted by Awsom Guy View Post
    Find F(x) given f(x) is:
    (x+2)(x^2`+ 4x + 3)^5
    Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
    My attempt:
    u=(x^2 + 4x + 3)
    u'=2x+4
    u^5
    u^6/6 + C
    (x^2 + 4x + 3)^6/6 + C
    I should have 12 instead of 6 on the bottom.
    Answer should be:
    (x^2 + 4x + 3)^6/12 + C
    Thank you
    \int{(x + 2)(x^2 + 4x + 3)^5\,dx} = \frac{1}{2}\int{2(x + 2)(x^2 + 4x + 3)^5\,dx}.

    Let u = x^2 + 4x + 3 so that \frac{du}{dx} = 2x + 4 = 2(x + 2)

    So the integral becomes

    \frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}

     = \frac{1}{2}\int{u^5\,du}

     = \frac{1}{2}\left[\frac{1}{6}u^6\right] + C

     = \frac{1}{12}u^6 + C

     = \frac{1}{12}(x^2 + 4x + 3) + C.
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  7. #7
    Member mathemagister's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    I know that but how do I do it properly wait I'll try and then give you a buzz.
    Remember the key is noticing that the derivative of the inside is (2x+4) which is exactly 2 times too big. (the outside is x+2)
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  8. #8
    Member mathemagister's Avatar
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    Quote Originally Posted by Prove It View Post
    \int{(x + 2)(x^2 + 4x + 3)^5\,dx} = \frac{1}{2}\int{2(x + 2)(x^2 + 4x + 3)^5\,dx}.

    Let u = x^2 + 4x + 3 so that \frac{du}{dx} = 2x + 4 = 2(x + 2)

    So the integral becomes

    \frac{1}{2}\int{u^5\,\frac{du}{dx}\,dx}

     = \frac{1}{2}\int{u^5\,du}

     = \frac{1}{2}\left[\frac{1}{6}u^6\right] + C

     = \frac{1}{12}u^6 + C

     = \frac{1}{12}(x^2 + 4x + 3) + C.
    Don't ruin it for him. At least put a spoiler around.

    This is part of the MHF terms of agreement:
    "When you see that someone is already helping a member and is waiting for the OP to give feedback, please don't give a full solution. "
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  9. #9
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    Quote Originally Posted by Awsom Guy View Post
    Find F(x) given f(x) is:
    (x+2)(x^2`+ 4x + 3)^5
    Well my problem is what do I do with the brackets. I tried but I can't get the answer right.
    My attempt:
    u=(x^2 + 4x + 3)
    u'=2x+4
    u^5
    u^6/6 + C
    (x^2 + 4x + 3)^6/6 + C
    Part of your problem is that you don't work "cleanly". For example, you say "u'= 2x+4" but you don't say why you need to know that or what you are doing with it. The very next line is "u^5" which is meaningless. I suspect that if you had written out the entire integral you would have seen where you went wrong.

    u= x^2+ 4x+ 3 so u'= du/dx= 2x+ 4 and du= (2x+4)dx. The whole point of taking the derivative there is to be able to decide how to replace "dx" in the integral. The crucial point is that du= (2x+4)dx but your integral has, in addition to (x^2+ 4x+ 3)^5, (x+2)dx, not (2x+4)dx.

    There are two paths people use. From your du= (2x+4)dx, you can factor a "2" out and have du= 2(x+2)dx so du/2= (x+2)dx.

    Or you can write dx= du/(2x+4) and replace that: (x+2)dx= (x+2)/(2x+4) du= (x+2)/(2(x+2)) du= (1/2)du.

    Either way, your integral is now \frac{1}{2}\int u^5 du, not just \int u^5 du.

    I should have 12 instead of 6 on the bottom.
    Answer should be:
    (x^2 + 4x + 3)^6/12 + C
    Thank you
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  10. #10
    Member Awsom Guy's Avatar
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    Thanks heaps prove it, this is how I learn.
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  11. #11
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    Quote Originally Posted by mathemagister View Post
    Don't ruin it for him. At least put a spoiler around.

    This is part of the MHF terms of agreement:
    "When you see that someone is already helping a member and is waiting for the OP to give feedback, please don't give a full solution. "
    Some people, like the OP for example, learn by having a solution to refer back to to check the answer they have come up with. The OP said he would be back after trying it himself.
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  12. #12
    Member Awsom Guy's Avatar
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    Thanks guys, I realised my problem I forgot the half in the front. Or else i would have gottn it thanks guys.
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  13. #13
    Member mathemagister's Avatar
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    Quote Originally Posted by Prove It View Post
    Some people, like the OP for example, learn by having a solution to refer back to to check the answer they have come up with. The OP said he would be back after trying it himself.
    I know, but I was just saying, because that got me a warning before.

    And Awsom Guy, great that you understand it now
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  14. #14
    Member Awsom Guy's Avatar
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    Oh, thanks anyways I knew I needed a half in there somehow but i wasn;t sure how to get it thanks everyone.
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