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Math Help - Another homework problem: Calculus

  1. #1
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    Another homework problem: Calculus

    Suppose f, f', f", are continuous on [0, ln2] and that f(0) = 0, f'(0) = 3, f(ln2) = 6, f'(ln2) = 4 and the integral of (e^-2x * f(x) dx from 0 to ln 2) = 3.


    Find the vlaue of the integral of (e^-2x *f"(x) dx) from 0 to ln 2.
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  2. #2
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    Hello, Susie!

    Suppose f,\;f',\'f'' are continuous on [0,\,\ln2]

    and that f(0) = 0,\;f'(0) = 3,\;f(\ln2) = 6,\;f'(\ln2) = 4

    and:  \int^{\ln2}_0\!\!e^{-2x}f(x)\,dx\;= \;3

    Find the value of: I \;= \;\int^{\ln 2}_0 e^{-2x}\cdot f''(x)\,dx
    We will integrate by parts . . .

    . . Let: u = e^{-2x}\qquad\qquad dv = f''(x)\,dx

    . . Then: du = -2e^{-2x}dx\quad v = f'(x)

    And we have: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\int e^{-2x}\!\cdot\!f'(x)\,dx


    Do by-parts again:

    Let: u = e^{-2x}\qquad\qquad dv = f'(x)\,dx

    Then: du = -2e^{-2x}dx\quad v = f(x)

    So we have: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\left[e^{-2x}\!\cdot\!f(x) + 2\int e^{-2x}\!\cdot\!f(x)\,dx\right]

    and: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx


    Now we evaluate from x = 0 to x = \ln 2
    . . Note that: e^{-2\cdot\ln 2}\:=\:e^{\ln(2^{-2})} \:= \:e^{\ln\frac{1}{4}} \;=\;\frac{1}{4}

    We have: I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l  n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}
    . . . . . . . . . . . . . . . . \downarrow . . . . . . . . . . . .. \downarrow . . . . . . . . . \downarrow
    Hence: . . I \;= \quad\;\frac{1}{4}(4) \quad\;+ \quad\;2\left(\frac{1}{4}\right)(6) \quad + \qquad4\;(3)

    . . . . . . . I \;= \; 1 + 3 + 12 \;= \;16


    But someone check my work . . . please!
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, Susie!


    We will integrate by parts . . .

    . . Let: u = e^{-2x}\qquad\qquad dv = f''(x)\,dx

    . . Then: du = -2e^{-2x}dx\quad v = f'(x)

    And we have: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\int e^{-2x}\!\cdot\!f'(x)\,dx

    Do by-parts again:

    Let: u = e^{-2x}\qquad\qquad dv = f'(x)\,dx

    Then: du = -2e^{-2x}dx\quad v = f(x)

    So we have: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\left[e^{-2x}\!\cdot\!f(x) + 2\int e^{-2x}\!\cdot\!f(x)\,dx\right]

    and: I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx


    Now we evaluate from x = 0 to x = \ln 2
    . . Note that: e^{-2\cdot\ln 2}\:=\:e^{\ln(2^{-2})} \:= \:e^{\ln\frac{1}{4}} \;=\;\frac{1}{4}

    We have: I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l  n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}
    . . . . . . . . . . . . . . . . \downarrow . . . . . . . . . . . .. \downarrow . . . . . . . . . \downarrow
    Hence: . . I \;= \quad\;\frac{1}{4}(4) \quad\;+ \quad\;2\left(\frac{1}{4}\right)(6) \quad + \qquad4\;(3)

    . . . . . . . I \;= \; 1 + 3 + 12 \;= \;16


    But someone check my work . . . please!
    I think that you have substitutedx=ln2 but forgor to substitute x=0
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by malaygoel
    I think that you have substitutedx=ln2 but forgor to substitute x=0
    I think the answer may be 16-3/4=61/4
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  5. #5
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    Hello, malaygoel!

    I think that you have substituted x = \ln2 but forgot to substitute x = 0
    You're right . . . *blush*
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, malaygoel!


    You're right . . . *blush*
    Could you please explain the solution given in the thread "Calculus Problem"
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  7. #7
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    how did you get 16- 3/4? I got 16 - 3.

    Can you explain your answer?
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx


    We have: I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l  n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}
    sorobsn has substituted only ln2
    and got the answer 16
    can you follow upto here
    if we substitute 0 we will get 1/3
    result of integration is difference of the resulting function at upper limit and lower limit
    tell me if I am not clear to you
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  9. #9
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    sorry I don't see how still.
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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    sorry I don't see how still.
    you got 16 right
    explain me how you got 3.
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  11. #11
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    I substituted 0 in for x, this is after substituting the ln 2

    -(e^-2(0)f'(0) + 2e^-2(0)f(0))

    -(1(3) + 2(1)(0)) = -3
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    I substituted 0 in for x, this is after substituting the ln 2

    -(e^-2(0)f'(0) + 2e^-2(0)f(0))

    -(1(3) + 2(1)(0)) = -3
    thank you for correcting me
    I was wrong
    it is 16 -3 =13
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