If f(x) = int_{x^2}^{x^2 + 1} e^{-t^2} dt,the | vedclass

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(D) To find where $f(x)$ increases,we calculate the derivative $f'(x)$ using Leibniz's rule:$f'(x) = \frac{d}{dx} \int_{x^2}^{x^2 + 1} e^{-t^2} dt = e

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$(-\infty, 0)$

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(D) To find where $f(x)$ increases,we calculate the derivative $f'(x)$ using Leibniz's rule:$f'(x) = \frac{d}{dx} \int_{x^2}^{x^2 + 1} e^{-t^2} dt = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$$f'(x) = e^{-(x^4 + 2x^2 + 1)} \cdot (2x) - e^{-x^4} \cdot (2x)$$f'(x) = 2x e^{-x^4} \left( e^{-(2x^2 + 1)} - 1 \right)$Since $e^{-(2x^2 + 1)} For $f'(x) > 0$,we need $2x$ to be negative,which implies $x Thus,$f(x)$ increases in $(-\infty, 0)$.

If $f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} dt$,then $f(x)$ increases in

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