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Math Help - Question about Integrals

  1. #1
    Member FalconPUNCH!'s Avatar
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    Question about Integrals

    Do you only change the limits of integration when substituting? I get confused and don't know when to change the limits of integration.
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Do you only change the limits of integration when substituting? I get confused and don't know when to change the limits of integration.
    You change the limit of an integral when you have a change of variables.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by polymerase View Post
    You change the limit of an integral when you have a change of variables.
    Can you give me some examples?
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Can you give me some examples?
    Sure.

    Find \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx

    Now, in order to find this, you must first solve the intergrand. So one way to solve this question is to use the technique of substitution.

    So I'm going to let u=x^2+4x+5. I have introduced a new variable u in place of x. In the orginal function, the limit was x from 2 to 4 but now my intergral is in terms of u, which MAY OR MAY NOT go from 2 to 4. Thus to find out what it does go to, you plug in x=2 into the equation u=x^2+4x+5 to find out what u goes to when x goes to 2. And you do the same for the other limit, 4. In the end, you should find the new limits of u to be from 17 to 37.

    Thus, \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx=\int^{37}_{17} \frac{du}{u}

    You then easily find the answer to be \ln \left(\frac{37}{17}\right) since \int^{37}_{17} \frac{du}{u}=\ln|u||^{37}_{17}. You can also after solving the integral, re-substitute then solve again in terms of x from 2 to 4. Same answer!

    Hope this helps!
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by polymerase View Post
    Sure.

    Find \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx

    Now, in order to find this, you must first solve the intergrand. So one way to solve this question is to use the technique of substitution.

    So I'm going to let u=x^2+4x+5. I have introduced a new variable u in place of x. In the orginal function, the limit was x from 2 to 4 but now my intergral is in terms of u, which MAY OR MAY NOT go from 2 to 4. Thus to find out what it does go to, you plug in x=2 into the equation u=x^2+4x+5 to find out what u goes to when x goes to 2. And you do the same for the other limit, 4. In the end, you should find the new limits of u to be from 17 to 37.

    Thus, \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx=\int^{37}_{17} \frac{du}{u}

    You then easily find the answer to be \ln \left(\frac{37}{17}\right) since \int^{37}_{17} \frac{du}{u}=\ln|u||^{37}_{17}. You can also after solving the integral, re-substitute then solve again in terms of x from 2 to 4. Same answer!

    Hope this helps!
    Oh so I have been doing them right I wasn't sure but now I get it. Thank You
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