Let g be the function given by $\displaystyle g(x) = \int_{0}^{x}sin(t^2)dt$ for -1 < x < 3. On which interval is g decreasing?
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Originally Posted by xxlvh Let g be the function given by $\displaystyle g(x) = \int_{0}^{x}sin(t^2)dt$ for -1 < x < 3. On which interval is g decreasing? Notice that $\displaystyle g'(x) = \sin (x^2)$. Now you need to find $\displaystyle g'>0$?
Ok, I solved it by graphing and the solution I had was approximately 1.772 < x < 2.507
Originally Posted by xxlvh 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: $\displaystyle \sin (y) > 0 $ for what $\displaystyle y$? Now set $\displaystyle y=x^2$ and solve for $\displaystyle x$.
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