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Math Help - hi plz help in this integral homework?????????

  1. #1
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    hi plz help in this integral homework?????????

    hello all
    if u can plz help me with this homework i attached it here

    thx 4 all
    Attached Thumbnails Attached Thumbnails hi plz help in this integral homework?????????-clip_image002.jpg  
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eman.kh View Post
    hello all
    if u can plz help me with this homework i attached it here

    thx 4 all
    f(x) = 6x - e^{1 - 2x}

    a)
    A = \int_1^5 (6x - e^{1 - 2x})dx
    and
    W = \frac{1}{4 - 1} \int_1^4 (6x - e^{1 - 2x})dx

    To do either of these we need
    \int (6x - e^{1 - 2x})dx

    Let's rewrite this a little:
    = \int (6x - e^1 \cdot e^{-2x})dx

    = \int 6x~dx - e \int e^{-2x}~dx

    The first integral is elementary, for the second one use the substitution u = -2x and the fact that \int e^u ~ du = e^u + C.

    b)
    f^{\prime}(x) = 6 - e^{1 - 2x} \cdot -2 = 6 + 2e^{1 - 2x}

    f^{\prime \prime}(x) = 2e^{1 - 2x} \cdot -2 = -4e^{1 - 2x}

    So just plug these into the expression f^{\prime \prime} + 2f^{\prime} - 12 and show that it has a value of 0.

    c)
    Concavity is determined by the second derivative. So to get the inflection points solve f^{\prime \prime}(x) = 0 for x. (Hint: e^{x} \geq 0 for all x, so are there any values of x such that the second derivative is 0?)

    To determine the concavity of the function, if f^{\prime \prime}(x) > 0 the function is concave up, if f^{\prime \prime}(x) < 0 the function is concave down.

    -Dan
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  3. #3
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    really thanks alot 4 helping me,idon't know how to thank u .

    thanks.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eman.kh View Post
    really thanks alot 4 helping me,idon't know how to thank u .

    thanks.
    That is thanks enough for me!

    -Dan
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