Past exam of the mathematical methods II course of the University of Oxford / 2017 / 5 / Solution

Let $G(s)={\int }_0^s{e}^{t^2}dt$. The fundamental theorem of calculus gives $G^{\prime }(s)=e^{s^2}$. Applying the chain rule to $F(x,y)=G(x{y}^2)$ givesNo elementary antiderivative of $e^{t^2}$ is needed: differentiation of the variable upper limit supplies the integrand directly.