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Thread: 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. $\displaystyle m(b-a) <= \int^b_a f(x)dx <= M(b-a)$

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

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

    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
    }
    $\displaystyle g(x) = \int ^x_0 f(t)dt$

    (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. $\displaystyle m(b-a) <= \int^b_a f(x)dx <= M(b-a)$

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

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

    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
    }
    $\displaystyle g(x) = \int ^x_0 f(t)dt$

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

    $\displaystyle 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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