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Math Help - Finding the derivative of this integral using part I AND part II of FTC

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    Finding the derivative of this integral using part I AND part II of FTC

    I have to find the derivative of the same integral twice, once using each part of the FTC. The integral is from sin(x) to 2 of the function (3^t+7). I think using part 1 it would just be 3^2+7 (but that seems way too simple?) and I don't really know how to do it using part two. I know part two deals with antiderivatives, but I don't know how that helps me here.
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    Quote Originally Posted by fattydq View Post
    I have to find the derivative of the same integral twice, once using each part of the FTC. The integral is from sin(x) to 2 of the function (3^t+7).
    \int_{sin(x)}^2 3^t+ 7 dt? That would be a function of x, not a number. Are you sure you have given the problem correctly?

    I think using part 1 it would just be 3^2+7 (but that seems way too simple?) and I don't really know how to do it using part two. I know part two deals with antiderivatives, but I don't know how that helps me here.
    Last edited by mr fantastic; February 12th 2009 at 04:47 PM. Reason: Fixed quote tags
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    Of course I gave the problem correctly. It says consider the given function h(x)=integral from sin(x) to 2 of (3^t+7)dt and then #1 is find h prime of x using part 1 of the FTC and #2 is the same just using part II

    using part 1 I think it would be -(3^sin(x)+7) times cos(x). Does this sound correct? And how would I go about using part II to find the same answer?
    Last edited by mr fantastic; February 12th 2009 at 04:59 PM. Reason: Merged posts
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    Quote Originally Posted by fattydq View Post
    using part 1 I think it would be -(3^sin(x)+7) times cos(x). Does this sound correct? Mr F says:Yes.

    And how would I go about using part II to find the same answer?
    Now calculate the integral and differentiate the result. Note that \int 3^t \, dt = \int e^{(\ln 3) t} \, dt \, ....
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    Quote Originally Posted by mr fantastic View Post
    Now calculate the integral and differentiate the result. Note that \int 3^t \, dt = \int e^{(\ln 3) t} \, dt \, ....
    what do you mean by calculate the integral? Just evaluate it?
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    Quote Originally Posted by fattydq View Post
    what do you mean by calculate the integral? Just evaluate it?
    Well yes. That's why I gave a result that will help you ....
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    I'm sorry I just don't understand. Could you go step by step?

    I just don't see how you can evaluate a function with an exponent of sinx when one of the values you have to plug in is sin(x)
    Last edited by mr fantastic; February 12th 2009 at 05:18 PM. Reason: Merged posts
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    Quote Originally Posted by fattydq View Post
    I'm sorry I just don't understand. Could you go step by step?

    I just don't see how you can evaluate a function with an exponent of sinx when one of the values you have to plug in is sin(x)
    \left[ \frac{1}{\ln 3} \, e^{(\ln 3) t} + 7t \right]_{\sin x}^2 = \left[ \frac{1}{\ln 3} \, 3^t + 7t \right]_{\sin x}^2 = \, ....

    where I'm assuming you know what to do with integral terminals.

    Now differentiate the result.
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    yes but I'm saying how can I put sin(x) in for t which is an exponent, when I don't know what x is. It's just sin(x) which doesn't help me at all.

    It's like you're speaking a foreign language man. I know that in the end my answer will be the same as in part I I just don't understand how to arrive to that, and what you're saying is just jumbling tons of numbers into the mix, I don't understand how that will lead me to the original answer I got when using part I.
    Last edited by mr fantastic; February 12th 2009 at 05:28 PM. Reason: Merged posts
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    Quote Originally Posted by mr fantastic View Post
    \left[ \frac{1}{\ln 3} \, e^{(\ln 3) t} + 7t \right]_{\sin x}^2 = \left[ \frac{1}{\ln 3} \, 3^t + 7t \right]_{\sin x}^2 = \, ....

    where I'm assuming you know what to do with integral terminals.

    Now differentiate the result.
    *Sigh* This is what you should have been taught to do with integral terminals:

     \left( \frac{1}{\ln 3} \, 3^2 + 7(2) \right) - \left( \frac{1}{\ln 3} \, 3^{\sin x} + 7(\sin x) \right)

    Simplify this. Now differentiate with respect to x.
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    Quote Originally Posted by fattydq View Post
    ... consider the given function h(x)=integral from sin(x) to 2 of (3^t+7)dt and then #1 is find h prime of x using part 1 of the FTC and #2 is the same just using part II
    here you go ...

    h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt

    h(x) = -\int_2^{\sin{x}} 3^t + 7 \, dt

    h'(x) = -(3^{\sin{x}} + 7) \cdot \cos{x}

    h'(x) = -3^{\sin{x}}\cos{x} - 7\cos{x}



    h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt

    h(x) = \left[\frac{3^t}{\ln{3}} + 7t\right]_{\sin{x}}^2

    h(x) = \left[\frac{3^2}{\ln{3}} + 7(2)\right] - \left[\frac{3^{\sin{x}}}{\ln{3}} + 7\sin{x}\right]

    h'(x) = 0 - \left[\frac{3^{\sin{x}}}{\ln{3}} \cdot \cos{x} \cdot \ln{3} + 7\cos{x}\right] = -3^{\sin{x}}\cos{x} - 7\cos{x}
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    Quote Originally Posted by mr fantastic View Post
    *Sigh* This is what you should have been taught to do with integral terminals:

     \left( \frac{1}{\ln 3} \, 3^2 + 7(2) \right) - \left( \frac{1}{\ln 3} \, 3^{\sin x} + 7(\sin x) \right)

    Simplify this. Now differentiate with respect to x.
    I appreciate you helping me out man, I'm sorry for being such a pain

    I understand that solving that will get me the answer I want but how did you get those terms to begin with?

    Imagine I'm starting from the beginning, with the original integral and solving it using part two...

    Quote Originally Posted by skeeter View Post
    here you go ...

    h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt

    h(x) = -\int_2^{\sin{x}} 3^t + 7 \, dt

    h'(x) = -(3^{\sin{x}} + 7) \cdot \cos{x}

    h'(x) = -3^{\sin{x}}\cos{x} - 7\cos{x}



    h(x) = \int_{\sin{x}}^2 3^t + 7 \, dt

    This. In between these two steps what are you guys doing to reach this next step?

    h(x) = \left[\frac{3^t}{\ln{3}} + 7t\right]_{\sin{x}}^2

    h(x) = \left[\frac{3^2}{\ln{3}} + 7(2)\right] - \left[\frac{3^{\sin{x}}}{\ln{3}} + 7\sin{x}\right]

    h'(x) = 0 - \left[\frac{3^{\sin{x}}}{\ln{3}} \cdot \cos{x} \cdot \ln{3} + 7\cos{x}\right] = -3^{\sin{x}}\cos{x} - 7\cos{x}
    I wrote in your quote box exactly where I get stuck
    Last edited by mr fantastic; February 13th 2009 at 04:24 AM. Reason: Merged posts, added red and replaced a tf with a hat ....
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    You have remember that the derivative of 3^t= (ln3)3^t so in order to integrate 3^t you need a ln3, thats where they get the antiderivative of 3^t/(ln3). To cancel the ln3 you get when you derive 3^t/(ln3).
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    Quote Originally Posted by aervid View Post
    You have remember that the derivative of 3^t= (ln3)3^t so in order to integrate 3^t you need a ln3, thats where they get the antiderivative of 3^t/(ln3). To cancel the ln3 you get when you derive 3^t/(ln3).
    wow yet another thing I wasn't aware I should know......calc II is ridiculous lol
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    Quote Originally Posted by fattydq View Post
    I appreciate you helping me out man, I'm sorry for being such a pain

    I understand that solving that will get me the answer I want but how did you get those terms to begin with?

    Imagine I'm starting from the beginning, with the original integral and solving it using part two...



    I wrote in your quote box exactly where I get stuck
    Both of us are doing nothing more than simple integration (the result I gave you way back is useful for doing this).
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