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Thread: Is there a correct/proper way to show the FTC?

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    Is there a correct/proper way to show the FTC?

    Is there a correct/proper way to show the FTC?-capture.png

    So, i know how to apply the FTC here directly. We just substitute 2x into sin(t^2+2) for t and multiply by derivative of 2x. However, I want to be able to show my work and go through the steps? Is there a way to do that cause im kinda confused on how to.

    Something along the lines of:

    Let G(x) be the antiderivative of sin(t^2+2) then G'(x) = sin(t^2+2)
    Im kinda confused what to do after to show the notation?
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    Re: Is there a correct/proper way to show the FTC?

    Would this be correct?
    Let f(x) = antiderivative of the sin(x^2+2)
    f'(x) = sin(x^2+2)
    f'(2x) = sin((2x)^2+2)*2
    Last edited by lc99; Dec 17th 2017 at 09:33 AM.
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    Re: Is there a correct/proper way to show the FTC?

    Quote Originally Posted by lc99 View Post
    Would this be correct?
    Let f(x) = antiderivative of the sin(x^2+2)
    f'(x) = sin(x^2+2)
    f'(2x) = sin((2x)^2+2)*2
    Suppose that each of $g~\&~h$ is a differentiable function and $\Phi (x) = \int_{g(x)}^{h(x)} {\varphi (t)dt}$
    then ${D_x}(\Phi (x)) = h'(x) \cdot \left( {\varphi \circ h(x)} \right) - g'(x) \cdot \left( {\varphi \circ g(x)} \right)$

    In this question, $h(x)=2x,~g(x)=1~\&~\varphi(t)-\sin(t^2+2)$
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    Re: Is there a correct/proper way to show the FTC?

    I'm kinda confused haha! I'm thinking that your way is more complex. I was justing wondering when i do these kind of problems like on my exams, for instance, is there a way to show my work? Like a simpler way? I'm afraid i might lose points if i just substitude the 2x into the sin(x) function . Is it okay to just write the answer without defining a function?
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    Re: Is there a correct/proper way to show the FTC?

    Quote Originally Posted by lc99 View Post
    I'm kinda confused haha! I'm thinking that your way is more complex. I was justing wondering when i do these kind of problems like on my exams, for instance, is there a way to show my work? Like a simpler way? I'm afraid i might lose points if i just substitude the 2x into the sin(x) function . Is it okay to just write the answer without defining a function?
    Plato's method isn't that bad when expressed in less advanced notation.

    what we're after is

    $\dfrac{d}{dx} \left(\displaystyle \int \limits_{f(x)}^{g(x)}~h(t)~dt \right)= h(g(x))\cdot g^\prime(x) - h(f(x))\cdot f^\prime(x)$

    here

    $g(x) = 2x,~g^\prime(x) = 2$

    $f(x) = 1,~f^\prime(x) = 0$

    $h(t) = \sin(t^2+2)$

    $\dfrac{d}{dx} \displaystyle \int \limits_{1}^{2x}~\sin(t^2+2)~dt = \sin((2x)^2)+2)\cdot 2 - \sin(1^2 + 2)\cdot 0 = 2 \sin(4x^2+2)$
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    Re: Is there a correct/proper way to show the FTC?

    More general than "FTC", Liebniz's theorem:
    $\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(t,x)dy= f(\beta(x), x)\frac{d\beta}{dx}- f(\alpha(x), x)\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(t, x)}{\partial x} dt$.
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    Re: Is there a correct/proper way to show the FTC?

    Quote Originally Posted by HallsofIvy View Post
    More general than "FTC", Liebniz's theorem:
    $\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(t,x)dy= f(\beta(x), x)\frac{d\beta}{dx}- f(\alpha(x), x)\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(t, x)}{\partial x} dt$.
    $dy$?
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