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Thread: integrationand differeniation help

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

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


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


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


    And also another qn. $\displaystyle \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
    $\displaystyle \int_1^4 g(x) dx = 5 $
    and $\displaystyle \int_2^4 g(x) = 2 $


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


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


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

    (ii) Note that $\displaystyle \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 $\displaystyle \int_4^1 g(x) - \sqrt{x} dx = - \int_1^4 g(x) \, dx + \int_1^4 x^{1/2} \, dx$.

    (ii) Note that $\displaystyle \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 $\displaystyle \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 $\displaystyle \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: $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

    $\displaystyle y = \ln \sin x$.

    Let $\displaystyle u = \sin x$.

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