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    We have already shown the following

    \frac{d}{dx}\bigint_a^x{f(t)} dt = f(x)

    Now, given that

    \bigint_0^x{f(t)}dt = \bigint_x^1{t^2 f(t)}dt + x - \frac{x^5}{5} + C

    I need to find f(x) and show that C= -2/15

    So far, I've got  f(x)=\frac{d}{dx}\displaystyle(\  bigint_0^1{t^2 f(t)}dt - \bigint_0^x{t^2 f(t)} dt\displaystyle) + 1 - x^4

    But now I'm stuck. Any hints on where I should be heading?
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    What STEP paper is it?
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    Je ne sais pas. II or III, I *think*
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    let p(t)=t^2f(t) then
    int {0,x} f(t)dt=int{1,x} p(t)dt +x-x^5/5+C
    int {0,x} f(t)dt=-int{x,1} p(t)dt +x-x^5/5+C
    diff wrt to x
    f(x)=-p(x)+1-x^4............................. ..using the first result on the 2 integrals.
    i.e.
    f(x)=-x^2f(x)+(1+x^2)(1-x^2)
    (1+x^2)f(x)=(1+x^2)(1-x^2)
    f(x)=1-x^2

    putting back into the integrals

    lhs integral=int {0,x} 1-t^2 dt
    =x-x^3/3
    rhs integral=int{1,x} t^2-t^4 dt
    =[1/3-1/5]-[x^3/3-x^5/5]
    =2/15-x^3/3+x^5/5
    so lhs integral =rhs integral +x-x^5-2/15
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    by differentiating
    we have

    \frac{d}{dx}\bigint_0^x{f(t)}dt = \frac{d}{dx}\bigint_x^1{t^2 f(t)}dt + 1 - x^4

    then
    f(x)=-{x^2}f(x)+1-{x^4}

    so
    f(x)=1-x^2
    C=-2/15
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    Thank you very much
 
 
 
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