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Math Help - derivative of an integral

  1. #1
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    derivative of an integral

    Hi

    The goal is the find f'(x) if

    f(x) = Integral sign (a= sec x; b= x^2) (1 + 2u^2)^ (1/2) <- that is the square root of (1 + 2u^2)

    Hopefully that makes sense in how I am communicating the question.

    So I started with letting t= sec x so dx/dt = sec x tan x and I am thinking I need to do the same with the x^2 but I don't know if it should be a different letter and then what do I do???

    Any help or direction you can offer is greatly appreciated.

    This calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi

    The goal is the find f'(x) if

    f(x) = Integral sign (a= sec x; b= x^2) (1 + 2u^2)^ (1/2) <- that is the square root of (1 + 2u^2)

    Hopefully that makes sense in how I am communicating the question.

    So I started with letting t= sec x so dx/dt = sec x tan x and I am thinking I need to do the same with the x^2 but I don't know if it should be a different letter and then what do I do???

    Any help or direction you can offer is greatly appreciated.

    This calculus beginner
    The second fundemental thm of calculus is

    if F(x) = \int_a^x f(t)\, dt then F'(x) = f(x).

    For your problem, split the integral up. i.e.

     <br />
\int_{\sec x}^{x^2} \sqrt{1+ 2u^2}\, du = \int_{\sec x}^{a} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du <br />
     <br />
= - \int_a^{\sec x} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du,<br />

    where a is a constant. Then use the chain rule on each piece.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Danny View Post
    The second fundemental thm of calculus is

    if F(x) = \int_a^x f(t)\, dt then F'(x) = f(x).

    For your problem, split the integral up. i.e.

     <br />
\int_{\sec x}^{x^2} \sqrt{1+ 2u^2}\, du = \int_{\sec x}^{a} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du <br />
     <br />
= - \int_a^{\sec x} \sqrt{1+ 2u^2}\, du + \int_{a}^{x^2} \sqrt{1+ 2u^2}\, du,<br />

    where a is a constant. Then use the chain rule on each piece.
    It seems unnecessary to split up the integral. Assume that \int\sqrt{1+2u^2}\,du = F(u). That is to say, F'(u)=\sqrt{1+2u^2}.

    Thus \int_{\sec x}^{x^2}\sqrt{1+2u^2}\,du = F(x^2)-F(\sec x).

    Taking the derivative of this expression (using the chain rule) gives you \sqrt{1+2(x^2)^2}\cdot2x-\sqrt{1+2(\sec x)^2}\cdot\sec x\tan x.

    Then you can clean that up a bit if you want.
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    It seems unnecessary to split up the integral. Assume that \int\sqrt{1+2u^2}\,du = F(u). That is to say, F'(u)=\sqrt{1+2u^2}.

    Thus \int_{\sec x}^{x^2}\sqrt{1+2u^2}\,du = F(x^2)-F(\sec x).

    Taking the derivative of this expression (using the chain rule) gives you \sqrt{1+2(x^2)^2}\cdot2x-\sqrt{1+2(\sec x)^2}\cdot\sec x\tan x.

    Then you can clean that up a bit if you want.
    I believe what you did and what I'm suggesting is equivalent!
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  5. #5
    Super Member redsoxfan325's Avatar
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    There's no question that they give the same answer. It's just that your method involves an extra step - just one more chance to make a mistake.
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