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Math Help - Definite Intergration

  1. #1
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    Definite Intergration

    Hi
    i few question i am having trouble finding the correct answer:
    1) \int_0^\frac{\pi}{2}5x+sin2x dx

    i got \frac{5\pi^2}{8}+\frac{1}{2} however the answers says its \frac{5\pi^2}{8}+1

    2) \int_1^4(2+\frac{1}{x})^2 dx

    \frac{(2+\frac{1}{x})^3}{3}

    \frac{(2+\frac{1}{4})^3}{3}-\frac{(2+\frac{1}{1})^3}{3}

    which gives me \frac{33}{4} and this is incorrect.
    3) \int_0^1x^2(1-x) dx
    someone tell me how would i start this off.


    P.S
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  2. #2
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    Rockhard

    You can break the integral up separately as such

    \int_0^{\frac{\pi}{2}} 5x

    and

    \int_0^{\frac{\pi}{2}}sin(2X)

    you can calculate each definite integral then combine them using addition in the proper order the equation was in evaluating the first integral yields

    \int_0^{\frac{\pi}{2}} 5x = \frac{5x^2}{2}

    And the second, you can use a simple U-Substitution as such

    u = 2x
    du = 2 dx
    \frac{1}{2}du=dx

    Now integrating

    \frac{1}{2}\int_0^{\frac{\pi}{2}}sin(u) = -\frac{1}{2}cos(x)
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  3. #3
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    Quote Originally Posted by Paymemoney View Post


    3) \int_0^1x^2(1-x) dx
    someone tell me how would i start this off.
    Expand the bracket
     \int_0^1 x^2-x^3 dx

    =  x^3/3-x^4/4 |x=1, 0
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  4. #4
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    Perhaps if you would review some basic algebraic simplification, it would be of some good benefit in improving your integration skills, and more in depth into the concept of substitution
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    2) \int_1^4(2+\frac{1}{x})^2 dx

    \frac{(2+\frac{1}{x})^3}{3}

    \frac{(2+\frac{1}{4})^3}{3}-\frac{(2+\frac{1}{1})^3}{3}

    which gives me \frac{33}{4} and this is incorrect.
    Your integration is incorrect.

    \int_1^4 (2+\frac{1}{x})^2 ~ dx = \int_1^4 4 + \frac{4}{x} + \frac{1}{x^2} ~ dx = \int_1^4 4 ~ dx + 4\int_1^4 \frac{1}{x} ~ dx + \int_1^4 \frac{1}{x^2} ~ dx

    You may only use the rule that you used if you have an integral of the form \int x^n for n\neq -1.
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