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Math Help - Interval of Decrease

  1. #1
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    Interval of Decrease

    Let g be the function given by g(x) = \int_{0}^{x}sin(t^2)dt for -1 < x < 3. On which interval is g decreasing?
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  2. #2
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    Quote Originally Posted by xxlvh View Post
    Let g be the function given by g(x) = \int_{0}^{x}sin(t^2)dt for -1 < x < 3. On which interval is g decreasing?
    Notice that g'(x) = \sin (x^2).
    Now you need to find g'>0?
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  3. #3
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    Ok, I solved it by graphing and the solution I had was approximately 1.772 < x < 2.507
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  4. #4
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    Quote Originally Posted by xxlvh View Post
    Ok, I solved it by graphing and the solution I had was approximately 1.772 < x < 2.507
    Do you see how you get the number?

    Hint: \sin (y) > 0 for what y?
    Now set y=x^2 and solve for x.
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