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Math Help - Need help understanding Fundamental theorm

  1. #1
    Junior Member
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    Need help understanding Fundamental theorm

    Why is the Integral
    \frac{d}{dx}\int_1^{sinx} tant dt
    equal to

    tan(sinx) \cdot cosx

    Why don't you have to take the the antiderivative of tan(t). Like you would if it was
    \frac{d}{dx}\int_1^{sinx} t+5

    Also why is the lower limit not used when solving the problem?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello swatpup32
    Quote Originally Posted by swatpup32 View Post
    Why is the Integral
    \frac{d}{dx}\int_1^{sinx} tant dt
    equal to

    tan(sinx) \cdot cosx

    Why don't you have to take the the antiderivative of tan(t). Like you would if it was
    \frac{d}{dx}\int_1^{sinx} t+5

    Also why is the lower limit not used when solving the problem?
    It doesn't matter what the function of t is, the explanation goes like this:

    \frac{d}{dx}\int_1^{\sin x} f'(t)\,dt = \frac{d}{dx}\Big[f(t)\Big]_1^{\sin x}

    = \frac{d}{dx}\Big(f(\sin x) - f(1)\Big)

    = f'(\sin x) \times \frac{d}{dx}(\sin x) - 0

    =f'(\sin x) \cos x

    OK?

    Grandad
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  3. #3
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    Joined
    Apr 2009
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    Quote Originally Posted by swatpup32 View Post
    Why is the Integral
    \frac{d}{dx}\int_1^{sinx} tant dt
    equal to

    tan(sinx) \cdot cosx

    Why don't you have to take the the antiderivative of tan(t). Like you would if it was
    \frac{d}{dx}\int_1^{sinx} t+5

    Also why is the lower limit not used when solving the problem?
    You use the Fundamental Theorem of Calculus:
    <br />
\int_{a}^{x}f(t)dt=F(x)<br />
    and F'(x)=f(x)

    so you have
    <br />
(\int_{a}^{g(x)}f(t)dt)'=F(g(x))g'(x)<br />

    For the question with the lower limit, you have the formula, from Leibinz-Newton formula:
    <br />
\int_{a}^{b}f(x)dx=F(b)-F(a)=-(F(a)-F(b))=-\int_{b}^{a}f(x)dx<br />

    Hope you are familiar with Lagrange's notation.
    <br />
f'(x)=\frac{df}{dx}<br />
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