The triangle $ADB$ is right-angled at $A$, with $AB=\sqrt{2}$ and $AD=1/\sqrt{2}$. Another application of the Pythagorean theorem gives $DB=\sqrt{2+1/2}=\sqrt{5/2}$. At the angle with vertex $D$, the adjacent side is $AD$ and the hypotenuse is $DB$. Hence

The two equal legs meet at the right angle, so the Pythagorean theorem gives $A{B}^2+A{C}^2=B{C}^2=4$, with $AB=AC$. Thus $AC=\sqrt{2}$. The midpoint condition then yields

The normal vector direction is orthogonal to the tangent line. For finite nonzero slopes, their product is $-1$. The tangent has slope one, so the normal line has

At a point of tangency, the derivative of the curve equals the slope one of the straight line. The chain rule gives $2\cos 2x=1$. Since $0\le 2x\le \pi$, the unique solution is $x=\pi /6$. The corresponding height is $\sin (\pi /3)=\sqrt{3}/2$. Substitution in the tangent line gives

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.

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.

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.

Two applications of integration by parts lower the exponent by two. First, the sine boundary term vanishes:For $n\ge 1$, integrate the last expression once more:Thus the recurrence relation isThe starting value is $I_0={\int }_0^{\pi }\cos xdx=0$. Successive applications give $I_1=-2\pi$ and $I_2=-4{\pi }^3+24\pi$, followed by

Use integration by substitution $u=\ln x$, $du=dx/x$. The bounds become $3$ and $4$, and the remaining denominator factorizes as $(u-1)(u-2)$. The partial fraction decomposition isConsequently the definite integral isAll logarithm arguments here are positive on the integration interval.