Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 20

Write $d=b-a$ and $h=d/n$. The required Riemann sum isUsing $\sum k=n(n+1)/2$ and the given sum of squares,An equivalent expression separates its limit and finite-$n$ corrections:ConsequentlyThe function is continuous, so the Riemann sums converge to its definite integral, the desired area under the parabola. Both corrections are positive when $0<a<b$, agreeing with the geometric fact that these right-endpoint rectangles overestimate an increasing curve.

Apply the Leibniz integral rule, including both moving boundaries and the explicit parameter in the integrand. The upper bound has derivative $-\sin x$, the lower bound has derivative $\cos x$, and the integrand has partial derivative $-t^4{e}^{-x{t}^4}$ with respect to $x$. ThereforeThe formula remains valid when the bounds coincide or occur in reversed order, by the oriented definite integral convention. The integrand and its partial derivative are continuous on every finite integration interval.

For the first definite integral, use integration by substitution $t=\tan x$, with $dx=dt/(1+t^2)$, ${\cos }^{2}x=1/(1+t^2)$ and ${\sin }^{2}x=t^2/(1+t^2)$. This givesFor the second calculation, let $D=p{\cos }^{2}x+q{\sin }^{2}x$. Since $p,q>0$, this denominator is bounded away from zero; on compact positive parameter ranges the differentiated integrands have uniform bounds. Thus differentiation under the integral sign is justified and givesAdding the two equations and using the Pythagorean trigonometric identity in the numerators yields $J=-I_p-I_q$, henceFor $p=q$, this becomes $\pi /(2p^2)$, agreeing with direct integration of the constant integrand.