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Math Help - integrationand differeniation help

  1. #1
    Member helloying's Avatar
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    integrationand differeniation help

     \int_1^4 g(x) dx = 5
    and \int_2^4 g(x) = 2


    evulate (i) \int_4^1 g(x) - \sqrt{x} dx


    and (ii) \int_1^2 g(x) dx


    And also another qn.  \frac{d}{dx} ln \sin x
    thanks for the help.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by helloying View Post
     \int_1^4 g(x) dx = 5
    and \int_2^4 g(x) = 2


    evulate (i) \int_4^1 g(x) - \sqrt{x} dx


    and (ii) \int_1^2 g(x) dx


    And also another qn.  \frac{d}{dx} ln \sin x
    thanks for the help.
    (i) Using some simple properties of the integral, you have \int_4^1 g(x) - \sqrt{x} dx = - \int_1^4 g(x) \, dx + \int_1^4 x^{1/2} \, dx.

    (ii) Note that \int_1^2 g(x) \, dx = \int_1^4 g(x) \, dx - \int_2^4 g(x) \, dx .

    For the final question, use the chain rule.
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  3. #3
    Member helloying's Avatar
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    Quote Originally Posted by mr fantastic View Post
    (i) Using some simple properties of the integral, you have \int_4^1 g(x) - \sqrt{x} dx = - \int_1^4 g(x) \, dx + \int_1^4 x^{1/2} \, dx.

    (ii) Note that \int_1^2 g(x) \, dx = \int_1^4 g(x) \, dx - \int_2^4 g(x) \, dx .

    For the final question, use the chain rule.

    thank you. but for the last qn, i dont know what is a chain rule. how i would attemp this qn would be: because \frac{d}{dx}ln x = 1/x , then the ans for the qn will be 1/sinx . but i check the book my ans is wrong.
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  4. #4
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    Quote Originally Posted by helloying View Post
    thank you. but for the last qn, i dont know what is a chain rule. how i would attemp this qn would be: because \frac{d}{dx}ln x = 1/x , then the ans for the qn will be 1/sinx . but i check the book my ans is wrong.
    I'm not sure why you would be attempting this question if you don't know the chain rule.

    Chain rule: \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

    y = \ln \sin x.

    Let u = \sin x.

    Then y = \ln u \Rightarrow \frac{dy}{du} = \frac{1}{u} = \frac{1}{\sin x} and \frac{du}{dx} = \cos x.
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