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

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(A) Given $f(x) = \int_x^{x+1} e^{-t^2} dt$.Using Leibniz's rule for differentiation under the integral sign, we have:$f'(x) = e^{-(x+1)^2} \cdot \fra

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$\left(-\frac{1}{2}, \infty\right)$

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(A) Given $f(x) = \int_x^{x+1} e^{-t^2} dt$.Using Leibniz's rule for differentiation under the integral sign, we have:$f'(x) = e^{-(x+1)^2} \cdot \frac{d}{dx}(x+1) - e^{-x^2} \cdot \frac{d}{dx}(x)$$f'(x) = e^{-(x+1)^2} - e^{-x^2}$For $f(x)$ to be a decreasing function, we must have $f'(x) $e^{-(x+1)^2} - e^{-x^2} $e^{-(x+1)^2} Since the exponential function $e^u$ is strictly increasing, we have:$-(x+1)^2 Multiplying by $-1$ reverses the inequality:$(x+1)^2 > x^2$$x^2 + 2x + 1 > x^2$$2x + 1 > 0$$2x > -1$$x > -\frac{1}{2}$Thus, the interval in which $f(x)$ is decreasing is $\left(-\frac{1}{2}, \infty\right)$.

If $f(x) = \int_x^{x+1} e^{-t^2} dt$, then the interval in which $f(x)$ is decreasing is

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