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Math Help - Integral Help (Check if I'm doing these right)

  1. #1
    Member FalconPUNCH!'s Avatar
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    Integral Help (Check if I'm doing these right)

    Use the form of the definition of the integral given in the Properties of Integrals to evaluate integrals below:

    1.
     \int\limits_1^4{(x^2+2x-5)dx}

    I then get:

     \int\limits_1^4{(x^2)dx} +  \int\limits_1^4{(2x)dx} -  \int\limits_1^4{(5)dx}

    I do the calculations and end up with

     \frac{n(n+1)(2n+1)}{n^3} (\frac{27}{6})

    Take up the limit as n -> infinite and get

     \int\limits_1^4{(x^2+2x-5)dx} = \frac{-1}{2}

    2.  \int\limits_0^5{(1+2x^3)}dx

    basically I did the same thing for this problem and end up with

    \int\limits_0^5{(1+2x^3)}dx = 1


    Here's another problem that I quite don't understand.


    3. Find the Riemann sum for f(x) = sinx ,  0 \le x \le \frac{3pi}{2}
    with six terms, taking he saple points to be right endpoints. Give your answer correct to six decimal places.

    There are several other parts to this particular question but I think I can manage to answer those if I find out how to do this part.
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  2. #2
    Eater of Worlds
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    No, I am sorry to say, you are not doing them right.

    What exactly are you trying to do?. I see a possible relationship to a Riemann sum there.

    \int_{1}^{4}{(x^{2}+2x-5)}dx=21

    {\Delta}x=\frac{4-1}{n}=\frac{3}{n}

    By the right endpoint method:

    x_{k}=1+\frac{3k}{n}

    f(x_{k}){\Delta}=\left[(1+\frac{3k}{n})^{2}+2(1+\frac{3k}{n})-5\right](\frac{3}{n})

    =\frac{27k^{2}}{n^{3}}+\frac{36k}{n^{2}}-\frac{6}{n}

    \frac{27}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{36}{n^{2  }}\sum_{k=1}^{n}k-\frac{6}{n}\sum_{k=1}^{n}1

    Now, use the identities you learned. I see you used one in your post(the sum of the squares). Then, take the limit as n\rightarrow{\infty}.
    You're shooting for 21, not -1/2.

    Try the same technique on this one, if that's what you need to do.

    \int_{0}^{5}(1+2x^{3})dx=\frac{635}{2}
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by galactus View Post
    No, I am sorry to say, you are not doing them right.

    What exactly are you trying to do?. I see a possible relationship to a Riemann sum there.

    \int_{1}^{4}{(x^{2}+2x-5)}dx=21

    {\Delta}x=\frac{4-1}{n}=\frac{3}{n}

    By the right endpoint method:

    x_{k}=1+\frac{3k}{n}

    f(x_{k}){\Delta}=\left[(1+\frac{3k}{n})^{2}+2(1+\frac{3k}{n})-5\right](\frac{3}{n})

    =\frac{27k^{2}}{n^{3}}+\frac{36k}{n^{2}}-\frac{6}{n}

    \frac{27}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{36}{n^{2  }}\sum_{k=1}^{n}k-\frac{6}{n}\sum_{k=1}^{n}1

    Now, use the identities you learned. I see you used one in your post(the sum of the squares). Then, take the limit as n\rightarrow{\infty}.
    You're shooting for 21, not -1/2.

    Try the same technique on this one, if that's what you need to do.

    \int_{0}^{5}(1+2x^{3})dx=\frac{635}{2}
    Hmm... I see what I have done wrong. Thank you for your help. I still need help with the last question though which I added while you were posting.

    when using K, K = any number in that interval right?
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  4. #4
    Eater of Worlds
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    Break sin(x) up into 6 partitions over the designated limits

    \frac{\frac{3\pi}{2}-0}{6}=\frac{\pi}{4}

    That's the width of each rectangle. Use that in your sine to find the heighths of the rectangles. The area of a rectangle is width times height.
    Add them up.

    sin(\frac{\pi}{4})+sin(\frac{\pi}{2})+......+sin(\  frac{3\pi}{2})

    Then multiply that by your width of \frac{\pi}{4}

    Here's a graph:
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    I'm having problems simplifying this:

     f(x_{i}) = \frac{5}{n} [1 + 2(1 + \frac{5i}{n})^3]

    any help?
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  6. #6
    Member FalconPUNCH!'s Avatar
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    f(x_{i}) = \frac{5}{n} [1 + 2(1 + \frac{5i}{n})^3]

     = \frac{5}{n} + \frac{10}{n}(1 + \frac{5i}{n})^3

    I get to that point but the cube, or shall I say the 1 +, is bothering me. Is there any way to simplify it or something?
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  7. #7
    Member FalconPUNCH!'s Avatar
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    This is my solution to the second one.

    {\Delta}{x} = {\frac{5}{n}} and  x_i = \frac{5i}{n}

    f(xi) = [ 1 + 2(\frac{5i}{n})^3] \frac{5}{n}

    I simplify and get

     = [\frac{15}{n} + \frac{625i^3}{n^4}

    Do the stuff with the sigma notation and get

     \frac{15}{n} + \frac{625}{2}(\frac{n(n+1)}{n4})^2

    I take the limit of that and get \frac{625}{2}

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  8. #8
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    Looks like you need to add 5.

    {\Delta}x=\frac{5}{n}

    Then you get:

    \left[1+2(\frac{5k}{n})^{3}\right](\frac{5}{n})=\frac{1250k^{3}}{n^{4}}+\frac{5}{n}

    \frac{1250}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{5}{n}\  sum_{k=1}^{n}1

    \frac{1250}{n^{4}}\cdot\frac{n^{2}(n+1)^{2}}{4}+5

    = \frac{625}{n}+\frac{625}{2n^{2}}+\frac{635}{2}

    Now, see the limit?.
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  9. #9
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by galactus View Post
    Looks like you need to add 5.

    {\Delta}x=\frac{5}{n}

    Then you get:

    \left[1+2(\frac{5k}{n})^{3}\right](\frac{5}{n})=\frac{1250k^{3}}{n^{4}}+\frac{5}{n}

    \frac{1250}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{5}{n}\  sum_{k=1}^{n}1

    \frac{1250}{n^{4}}\cdot\frac{n^{2}(n+1)^{2}}{4}+5

    = \frac{625}{n}+\frac{625}{2n^{2}}+\frac{635}{2}

    Now, see the limit?.
    hmm. yeah I got to go over and make sure I didn't make the same mistake with other problems. Thanks
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