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  1. MHB Apprentice

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    #1
    So the question is…Evaluate the following…
    $ \displaystyle \frac{d}{dx} \left(\int _1^{x^2} \cos(t^2) \, dt \right)$

    I thought i could use the FTC on this because it states…
    $ \displaystyle \frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$

    but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem

    Thanks
    Last edited by ZaidAlyafey; January 10th, 2014 at 09:58. Reason: Use LaTeX

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    #2
    Since the lower bound is still a constant, it doesn't change the method you use to evaluate this. It could be at 0,1,2 or any constant and still get the same result. You're going to integrate and get a constant and then take the derivative of that constant, so you end up with 0 regardless.

    Here you'll need to be careful to use the chain rule when applying the FTOC. What progress have you made so far?

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    #3
    Quote Originally Posted by ISITIEIW View Post
    So the question is…Evaluate the following…
    d/dx(integration of 1 to x^2 of cos(t^2)dt)

    I thought i could use the FTC on this because it states…
    d/dx(integration of 0 to x of f(t)dt)

    but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem

    Thanks
    We are given to evaluate:

    $ \displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt$

    Now, the anti-derivative form of the FTOC tells us:

    $ \displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=\frac{d}{dx}\left(F(h(x))-F(g(x)) \right)$

    On the right, applying the chain rule, we obtain:

    $ \displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=F'(h(x))h'(x)-F'(g(x))g'(x)$

    Since $ \displaystyle F'(x)=f(x)$, we may write:

    $ \displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f(h(x))h'(x)-f(g(x))g'(x)$

    Applying this to the given problem, there results:

    $ \displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=\cos\left(\left(x^2 \right)^2 \right)(2x)-\cos\left(\left(1 \right)^2 \right)(0)$

    Simplifying, we find:

    $ \displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=2x\cos\left(x^4\right)$

  5. زيد اليافعي
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    #4
    Quote Originally Posted by ISITIEIW View Post

    but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
    Let us evaluate $ \displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)$ where $ \displaystyle 0<a<x$

    Then since $ \displaystyle \int_0^x f(t)\, dt =\int_0^a f(t)\, dt+ \int_a^x f(t)\, dt $

    $ \displaystyle \int_a^x f(t)\, dt=-\int_0^a f(t)\, dt+\int_0^x f(t)\, dt $

    If we differentiate then since the first integral is independent of $x$ we have

    $ \displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)=\frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$

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    #5
    The "Leibniz formula", a generalization of the fundamental theorem of calculus, says
    $ \displaystyle \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x, \beta(x))- \frac{d\alpha}{dx}f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt$

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