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Math Help - urgent hw help on calculus

  1. #1
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    urgent hw help on calculus

    find a function f and a number a such that 6+ integral of t=x to t=a f(t)/t^2dt = 2square root x for all x>0.
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    Quote Originally Posted by twilightstr View Post
    find a function f and a number a such that 6+ integral of t=x to t=a f(t)/t^2dt = 2square root x for all x>0.
    6 + \int_x^a \frac{f(t)}{t^2} \, dt = 2\sqrt{x}

    hint ...

    \frac{d}{dx}\left[\int_a^x f(z) \, dz\right] = f(x)
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    still need help

    i still dont know what to do from there. srry
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    Quote Originally Posted by twilightstr View Post
    i still dont know what to do from there. srry
    Re-arrange: \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}.

    Differentiate both sides wrt x: \frac{f(x)}{x^2} = -\frac{1}{\sqrt{x}} \Rightarrow f(x) = -x^{3/2}.

    Substitute this expression into \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x} and solve for a.
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  5. #5
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    its actually -6+ 2rootx but no worries i still understand it

    thanks
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  6. #6
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    Quote Originally Posted by twilightstr View Post
    its actually -6+ 2rootx but no worries i still understand it

    thanks
    No it's not. Study my reply carefully - I also reversed the integral terminals (so that the Fundamental Theorem of Calculus could be used to differentiate the integral).
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  7. #7
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    in the given problem, the upper bound is x and the lower bound is a.
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    Quote Originally Posted by twilightstr View Post
    in the given problem, the upper bound is x and the lower bound is a.
    I do realise that.

    Are you familiar with the following property:

    \int_{\alpha}^{\beta} f(x) \, dx = - \int^{\alpha}_{\beta} f(x) \, dx.

    That's the property I used.

    If you're familiar with the Fundamental Theorem of Calculus (see post #2) then you will understand why the integral terminals in your question need to be reversed before the integral can be differentiated.
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  9. #9
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    the answer in the book is f(x)= x^3/2 and a=9
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