The points of extremum of int_0^{x^2} frac{t^ | vedclass

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(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$. To find the points of extremum,we calculate the derivative $f'(x)$ using the Leibniz Integral

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$0, \pm 1, \pm 2$

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(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$. To find the points of extremum,we calculate the derivative $f'(x)$ using the Leibniz Integral Rule: $f'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$. $f'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot (2x)$. $f'(x) = \frac{(x^2-1)(x^2-4)}{2 + e^{x^2}} \cdot (2x)$. $f'(x) = \frac{(x-1)(x+1)(x-2)(x+2)(2x)}{2 + e^{x^2}}$. Setting $f'(x) = 0$,we get $x = 0, 1, -1, 2, -2$. Thus,the points of extremum are $0, \pm 1, \pm 2$.

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

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