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Math Help - Integrals

  1. #1
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    Integrals

    Hey guys. Can anyone please help me out with these problems?







    Integrals-capture3.jpgIntegrals-capture.jpgIntegrals-capture2.jpg
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  2. #2
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    Capture 1 is correct.

    I can't read what your upper limit is in Capture 2, but you're on the right track with your thinking.

    \displaystyle \int{\frac{x}{x^2-1}\,dx} = \frac{1}{2}\int{\frac{2x}{x^2 - 1}\,dx}.

    Now make the substitution \displaystyle u = x^2-1 so that \displaystyle \frac{du}{dx} = 2x.


    For Capture 3, again, substitution is the way to go.

    \displaystyle \int{\cos{2x}\,dx} = \frac{1}{2}\int{2\cos{2x}\,dx}.

    Now make the substitution \displaystyle u = 2x so that \displaystyle \frac{du}{dx} = 2.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Capture 1 is correct.

    I can't read what your upper limit is in Capture 2, but you're on the right track with your thinking.

    \displaystyle \int{\frac{x}{x^2-1}\,dx} = \frac{1}{2}\int{\frac{2x}{x^2 - 1}\,dx}.

    Now make the substitution \displaystyle u = x^2-1 so that \displaystyle \frac{du}{dx} = 2x.


    For Capture 3, again, substitution is the way to go.

    \displaystyle \int{\cos{2x}\,dx} = \frac{1}{2}\int{2\cos{2x}\,dx}.

    Now make the substitution \displaystyle u = 2x so that \displaystyle \frac{du}{dx} = 2.
    For my second problem, my upper limit was square root of e+1.

    And I know this may seem like a dumb question, but when you move 1/2 behind the integration sign, how do you know when to do that? I realize it is there to cancel out the 2 that is placed in front of the function, but is it due to the aftermath of the substitution rule?
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  4. #4
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    Yes, because you need to have \displaystyle 2x as a factor, not \displaystyle x. In order to have an equivalent integral, whatever you multiply inside the integral has to be divided outside the integral. You can only do this for constants though.
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  5. #5
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    There are two slightly different ways to think about that.

    Your integral is \int \frac{x dx}{x^2- 1} and you want to use the substitution u= x^2- 1. Then \frac{du}{dx}= 2x so du= 2x dx

    Now either
    1) Recognize that you have x dx in the numerator but need 2x dx so multiply and divide by 2:
    \int \frac{2dx}{2(x^2- 1)}= \frac{1}{2}\int\frac{2xdx}{x^2- 1}= \frac{1}{2}\int\frac{du}{u}
    or
    2) Divide both sides of du= 2xdx by 2 to get \frac{1}{2}du= x dx so that
    \int \frac{x dx}{x^2- 1}= \int \frac{\frac{1}{2}du}{u}= \frac{1}{2}\int\frac{du}{u}.

    Note that only constants can be moved out of the integral like that. If the "x" had not been in the numerator to begin with, we could not have used that substitution.
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  6. #6
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    Sorry to post again, I just want to make sure if I actually understand this.
    Attached Thumbnails Attached Thumbnails Integrals-c2.jpg  
    Last edited by mr fantastic; December 18th 2010 at 08:22 PM.
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  7. #7
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    Quote Originally Posted by FullBox View Post
    Sorry to post again, I just want to make sure if I actually understand this.
    What you've done is inefficient and in some ways, wrong. When you make the substitution u = x^2 - 1 you should also substitute for the integral terminals and then evaluate the integral without ever going back to the old variable x. x = \sqrt{e + 1} \Rightarrow u = e and x = \sqrt{2} \Rightarrow u = 1. So the new integral is ....
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  8. #8
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    So are you saying that I should change the upper and lower limits with respect to u, and forget about the ones with respect to x?
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  9. #9
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    Quote Originally Posted by FullBox View Post
    So are you saying that I should change the upper and lower limits with respect to u, and forget about the ones with respect to x?
    Yes, by doing that you won't need to rewrite the integral in terms of \displaystyle x at the end, you can just substitute the values of \displaystyle u.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    What you've done is inefficient and in some ways, wrong. When you make the substitution u = x^2 - 1 you should also substitute for the integral terminals and then evaluate the integral without ever going back to the old variable x. x = \sqrt{e + 1} \Rightarrow u = e and x = \sqrt{2} \Rightarrow u = 1. So the new integral is ....
    Thank you Mr. Fantastic and Prove it. I think I understand it now. I have done the problem over again, and hopefully I have done it correctly this time!

    EDIT: TYPO! The answer is 1/2, I realized I made a small error on the bottom when I had 1/2(lne). I made it equal to 1, when in reality it is 1/2*1=1/2.



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    Last edited by FullBox; December 19th 2010 at 12:31 PM.
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  11. #11
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    For the 3rd problem, you have to know the anti-derivative of the trig function cos(x).

    The anti-derivative for cos(x) is sin(x), but in this case we have a cos(2x) which makes finding the anti-derivative a not any harder to be honest. The anti-derivative for cos(2x) is sin(2x)/2

    basically your just taking the coefficient of x (which is 2) and dividing it by the anti-derivative and this works in pretty much any trig function.

    so in any case say finding the anti-derivative of sin(Cx) would equal -cos(Cx)/C where C is represents the coefficient.
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  12. #12
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    Thanks! I redid that problem and I think I get it now!

    EDIT: Just one small complication though. I checked the answers in the back of the book, and it has the same answer as me only positive, while mine is negative. I'm not sure why though.

    EDIT [again]: Nevermind, I just realized I calculated cos's antiderivative incorrectly; I put -sin rather than positive sin.



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