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Math Help - Integral review

  1. #1
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    Integral review

    1) It is known that if m f(x) M for a x b, then the following property of integrals is true. m(b-a) <= \int^b_a f(x)dx <= M(b-a)

    Use this property to estimate the value of the given integral.

    ____ < \int^6_0 3xe^{-x}dx < ____

    Would I just have to find the minimum and maximum on that interval, then plug everything in?

    2)
    Code:
     f(x) = {
    0                   if x < 0
    x                    if 0 <= x <= 1
    2 - x          if 1 <= x <= 2
    0                    if x > 2
    }
    g(x) = \int ^x_0 f(t)dt

    (a) Find an expression for g(x) similar to the one for f(x).
    Give your answer in the form below.



    State where g(x) is discontinuous (if anywhere)

    I've done everything but H and K, which I think I need to decide if it's continuous or not. I thought K would just be the Integral of 0, but that's not right, and I'm not sure how to go about solving for H on this one.

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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    1) It is known that if m f(x) M for a x b, then the following property of integrals is true. m(b-a) <= \int^b_a f(x)dx <= M(b-a)

    Use this property to estimate the value of the given integral.

    ____ < \int^6_0 3xe^{-x}dx < ____

    Would I just have to find the minimum and maximum on that interval, then plug everything in?

    YES

    2)
    Code:
     f(x) = {
    0                   if x < 0
    x                    if 0 <= x <= 1
    2 - x          if 1 <= x <= 2
    0                    if x > 2
    }
    g(x) = \int ^x_0 f(t)dt

    (a) Find an expression for g(x) similar to the one for f(x).
    Give your answer in the form below.



    State where g(x) is discontinuous (if anywhere)

    I've done everything but H and K, which I think I need to decide if it's continuous or not. I thought K would just be the Integral of 0, but that's not right, and I'm not sure how to go about solving for H on this one.

    H,K=0. btw, what is the meaning of this:
    x if 0 <= x <= 1
    2 - x if 1 <= x <= 2

    if x=1, we can't tell whether f(x) will equal x or 2-x.
    see comments within the quote.
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  3. #3
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    Quote Originally Posted by vince View Post
    see comments within the quote.
    Apologies about the typo, it should be:

    x if 0 <= x <= 1
    2 - x if 1 < x <= 2


    I forgot to mention, I've already tried 0 for both H and K. Neither were correct.
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  4. #4
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    oh....oops.

    notice that when
    g(x) = \int ^x_0 f(t)dt is evaluated when x is between 1 & 2, it's an intergal from 0 to 2 so that in the instance

    g(x) = \int ^x_0 f(t)dt = \int ^1_0 f(t)dt + \int ^x_1 f(t)dt

    str8forward from here right
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