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Math Help - Use form of the definition of the integral given in theorem 4 to evaluate integral

  1. #1
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    Use form of the definition of the integral given in theorem 4 to evaluate integral

    Theorem 4 says that if f is integrable on [a,b], then

    \int_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^nf(x_i)\Delta{x}

    where \Delta{x}=\frac{b-a}{n} and x_i=a+i\Delta{x}

    My question is \int_{-1}^{5} (1+3x)dx

    I have \Delta{x}=\frac{6}{n} ; x_i=\frac{6i}{n}

    so \int_{-1}^{5} (1+3x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^n\left[1+\frac{18i}{n}\right]\cdot\frac{6}{n}=

    =\lim_{n \rightarrow \infty}\frac{6}{n}\left[\sum_{i=1}^n1+\frac{18}{n}\sum_{i=1}^ni\right]=

    =\lim_{n \rightarrow \infty}\frac{6}{n}\left[n+\frac{18}{n}\cdot\frac{n(n+1)}{2}\right]=

    =\lim_{n \rightarrow \infty}6+54\cdot\frac{n^2+n}{n^2}=60

    but by taking the antiderivative I get 42.
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  2. #2
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Quote Originally Posted by amthomasjr View Post
    Theorem 4 says that if f is integrable on [a,b], then

    \int_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^nf(x_i)\Delta{x}

    where \Delta{x}=\frac{b-a}{n} and x_i=a+i\Delta{x}

    My question is \int_{-1}^{5} (1+3x)dx

    I have \Delta{x}=\frac{6}{n} ; x_i=\frac{6i}{n}

    Your x_i is incorrect, it should be x_i=-1+i\Delta x
    Thanks from amthomasjr
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  3. #3
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Ah, I see. Forgive me for wasting your time.
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  4. #4
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Quote Originally Posted by amthomasjr View Post
    Ah, I see. Forgive me for wasting your time.
    You learn by asking. Sounds trite, but it's true. You wasted no one's time.

    -Dan
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  5. #5
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Quote Originally Posted by Plato View Post
    Your x_i is incorrect, it should be x_i=-1+i\Delta x
    =\lim_{n \rightarrow \infty}\frac{6}{n}\left[\sum_{i=1}^n-2+\frac{18}{n}\sum_{i=1}^ni\right]=

    Instead of getting 6 + 54, I get -12 + 54.

    Thanks
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Quote Originally Posted by amthomasjr View Post
    I get -12 + 54.

    But -12+54=42 which is correct.
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  7. #7
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    Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

    Yes, that is what I meant.
    Thank you
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