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Math Help - evaluating a definite integral

  1. #1
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    evaluating a definite integral

    How would I evaluate this integral using the 2nd Fundamental Theorem of Calculus? Thanks in advance.

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    Ahh I love these questions.

    Let \frac{2i}{n}=x. Thus dx=\frac{2}{n}.

    Now rewrite to an integral integral with this substitution.

    \int_{L} 1 +x +x^2dx

    Now for the bounds.

    For the first bound let i=0 and for the second let i=n. So the bounds are from 0 to 2.

    Got it?
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    Quote Originally Posted by Jameson View Post
    Ahh I love these questions.

    Let \frac{2i}{n}=x. Thus dx=\frac{2}{n}.

    Now rewrite to an integral integral with this substitution.

    \int_{L} 1 +x +x^2dx

    Now for the bounds.

    For the first bound let i=0 and for the second let i=n. So the bounds are from 0 to 2.

    Got it?
    Shouldn't that second bound be \infty?

    -Dan
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    I don't believe so. Why would it be inifinity? On these problems i=0, then i=n. Is there another way you do it?
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  5. #5
    TD!
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    For the integral, the limits of x are indeed 0 to 2.
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  6. #6
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    Yea that explanation makes sense Jameson, but I'm still not sure how this would be solved once reaching that point. How'd you determine the bounds were 0 and 2?
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    Quote Originally Posted by thedoge View Post
    How would I evaluate this integral using the 2nd Fundamental Theorem of Calculus? Thanks in advance.



    The standard form the the Riemann Ingegral is,
    \lim_{n\to \infty} \sum_{k=1}^n f(a+k\Delta x)\Delta x
    In this case,
    \Delta x=\frac{2}{n}
    Thus,
    b-a=2
    But there is not "a" term,
    a=0 thus, b=2.
    And the function is,
    f(x)=1+x+x^2

    Thus we need to find,
    \int_0^2 1+x+x^2 dx
    Which we can easily evaluate by Fundamental theorem because this function is continous on the closed interval.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Shouldn't that second bound be \infty?

    -Dan
    Momentary brain fart!

    -Dan
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