The function f(x) = int_{x^2}^{x^2+1} e^{-t^2 | vedclass

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(B) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$.Using Leibniz's rule for differentiation under the integral sign:$f'(x) = e^{-(x^2+1)^2} \cdot \frac

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$(-\infty, 0)$

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(B) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$.Using Leibniz's rule for differentiation under the integral sign:$f'(x) = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$$f'(x) = e^{-(x^2+1)^2} \cdot (2x) - e^{-x^4} \cdot (2x)$$f'(x) = 2x \left( e^{-(x^2+1)^2} - e^{-x^4} \right)$Since $(x^2+1)^2 > x^4$ for all $x$,it follows that $-(x^2+1)^2 Therefore,$e^{-(x^2+1)^2} For $f(x)$ to be an increasing function,we require $f'(x) \ge 0$.Since the term in the parenthesis is always negative,$f'(x) \ge 0$ only when $2x \le 0$,which means $x \le 0$.Thus,$f(x)$ is increasing in the interval $(-\infty, 0)$.

The function $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$ is an increasing function in the interval ...

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