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

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(D) Let $F(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$F'(x) = \

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All of these

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(D) Let $F(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$.Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:$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$For extremum points,we set $F'(x) = 0$:$\frac{(x^2 - 1)(x^2 - 4)}{2 + e^{x^2}} \cdot 2x = 0$This implies $x = 0$ or $x^2 = 1$ or $x^2 = 4$.Solving these,we get $x = 0$,$x = \pm 1$,and $x = \pm 2$.Since all these values are included in the options provided,the correct answer is $D$.

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

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