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Thread: Fundmental Theorem of Calculus problem #2

  1. #1
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    Fundmental Theorem of Calculus problem #2

    The question is: The function f is differentiable on (−∞, ∞). Determine f (2) in the following cases:

    $\displaystyle \int_0^{f(x)}{t^2}dx={x^2}{(1+x)}} $

    So I let $\displaystyle {u}={f(x)}$ ==>$\displaystyle \int_0^u{t^2}{dt}$ ==>
    $\displaystyle \frac{d}{du} \int_0^u{t^2}{dt}={(f(u))^2}$

    So $\displaystyle \frac{d}{dx} \int_0^u {t^2}{dt}={(f(u))^2} \frac{du}{dx}$ ==> $\displaystyle {(f(x))^2}{f'(x)}={2x}+{3x^2}$

    Assuming that my steps are correct (and I'm not certain that they are), where do I go from here? For instance, what is $\displaystyle {f'(x)}$?

    Many thanks for any suggestions for this problem!
    CP


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  2. #2
    MHF Contributor

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    Re: Fundmental Theorem of Calculus problem #2

    First, I suspect there is a typo in your statement of the problem. Your integrand is [tex]x^2[tex] but the integral with respect to x!
    I suspect you meant $\displaystyle \int_0^{x^2} t^2 dt$. Is that correct? Assuming it is-

    I'm not sure why you would differentiate. If you just go ahead and integrate $\displaystyle \int t^2 dt= \frac{1}{3}t^3+ C$ so that $\displaystyle \int_0^{f(x)} t^2 dt= \frac{1}{3} (f(x))^3$.

    So your equation is $\displaystyle \frac{1}{3}(f(x))^3= x^2(1+ x)$

    Solve that for f(x). (f(x) is just a number. This is exactly the same as solving $\displaystyle \frac{1}{3}y^3= x^2(1+ x)$ for x.)
    Thanks from CrispyPlanet
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