Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 64

$f$ is increasing on the interval $(-1,1)$

$f(x) = \int_{0}^{x}(1-t^2)e^{t^2}~dt$ Then: $f'(x) = (1-x^2)e^{x^2}$ $f$ is increasing when $f'(x) \gt 0$ $(1-x^2)e^{x^2} \gt 0$ Since $e^{x^2}$ is always positive, we can find the interval when $1-x^2 \gt 0$: $(1-x^2) \gt 0$ $x^2 \lt 1$ $-1 \lt x \lt 1$ Therefore: $f$ is increasing on the interval $(-1,1)$