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Math Help - Not too sure what to do

  1. #1
    Junior Member
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    Unhappy Not too sure what to do

    Compute integral of x*f'(x)*f(x) dx from a to b if

    the integral of f^2(x)dx from a to b equals 2 and f(a)=f(b)=0.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    Compute integral of x*f'(x)*f(x) dx from a to b if

    the integral of f^2(x)dx from a to b equals 2 and f(a)=f(b)=0.
    use integration by parts
    integral of {x}*{f'(x)*f(x)} dx from a to b
    ={x*f^2(x)/2 from a to b} - {integral of 1*f^2(x)/2 dx from a to b}
    ={b*f^2(b)- a*f^2(a)} -2/2
    =b*0 -a*0 -1
    = -1
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  3. #3
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    Hello, Nichelle14!

    Did you follow Malaygoel's solution?

    Compute: I \;= \;\int^b_a x\cdot f'(x)\cdot f(x)\,dx

    if \int^b_a [f(x)]^2 = 2 and f(a) = f(b) = 0
    Integrate by parts . . .

    Let: u = x\qquad\qquad dv = f(x)\cdot f'(x)\,dx

    Then: du = dx\qquad v = \frac{1}{2}[f(x)]^2


    We have: . I \;= \;\underbrace{\frac{1}{2}x\left[f(x)\right]^2\bigg|^b_a} - \frac{1}{2}\underbrace{\int^b_a[f(x)]^2\,dx}

    . . I \;= \;\overbrace{\frac{1}{2}(b)[f(b)]^2 - \frac{1}{2}(a)[f(a)]^2} - \frac{1}{2}[2]

    . . I \;= \;\frac{1}{2}(b)(0^2) - \frac{1}{2}(a)(0^2) - 1 \;\;= \;0 - 0 -1 \;\;= -1
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