Find approximately the perimeter of the ellipse (x²)/25+
Find approximately the perimeter of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{3}=1$, applying Simpson's formula with $2n=2$.
Explanation of Method
The Cartesian arc-length integrand for an ellipse blows up at the endpoint of the major axis (where the tangent is vertical), so it is safer and cleaner to switch to the parametric form $x=a\cos t,\ y=b\sin t$, whose arc-length integrand $\sqrt{a^2\sin^2t+b^2\cos^2t}$ stays perfectly finite over the whole quarter-period $t\in[0,\pi/2]$; Simpson's rule is then applied to that integral, and the perimeter follows from the fourfold symmetry of the ellipse.
Step 1: Ellipse parameters and symmetry
$a=5$, $b=\sqrt3\approx1.732$. By symmetry the full perimeter is $P=4L_1$, where $L_1$ is the arc length over the first quadrant.
Step 2: Parametric arc-length integrand
With $x=5\cos t,\ y=\sqrt3\sin t$: $\dfrac{dx}{dt}=-5\sin t,\ \dfrac{dy}{dt}=\sqrt3\cos t$, so
\[f(t)=\sqrt{25\sin^2t+3\cos^2t},\qquad L_1=\int_0^{\pi/2}f(t)\,dt.\]
Step 3: Simpson's formula, $2n=2$
$h=\pi/4$, nodes $t=0,\pi/4,\pi/2$:
\[f(0)=\sqrt3\approx1.732051,\quad f(\pi/4)=\sqrt{14}\approx3.741657,\quad f(\pi/2)=5.\]
\[S=1.732051+4(3.741657)+5=21.698680.\]
Step 4: Evaluate $L_1$
\[L_1=\frac{h}{3}S=\frac{\pi/4}{3}\times21.698680=\frac{\pi}{12}\times21.698680\approx5.680701.\]
Step 5: Perimeter
\[P=4L_1\approx4\times5.680701\approx22.722805.\]
\[\boxed{P\approx22.72\ \text{units}}\]
Step 6: Accuracy remark
The true perimeter (from the complete elliptic integral of the second kind) is $\approx22.415$, so this very coarse Simpson approximation ($2n=2$, only three sample points) overshoots by about $1.4\%$. That is expected: with so few subdivisions over an integrand with substantial curvature, the truncation error of the method itself dominates; using more subintervals would bring the estimate much closer to the true value, as the problem notes.
Result
Final answer:
\[\boxed{P\approx22.72\ \text{units}\ (2n{=}2,\ \text{coarse approximation})}\]