Using Simpson's rule, approximately calculate

Using Simpson's rule, approximately calculate the perimeter of the ellipse $\dfrac{x^2}{\sqrt8}+\dfrac{y^2}{\sqrt6}=1$.

The given equation has $x^2/a^2+y^2/b^2=1$ with $a^2=\sqrt8$ and $b^2=\sqrt6$, so $a=8^{1/4}\approx1.6818$ and $b=6^{1/4}\approx1.5651$.

Step 1: Parametrize and find the speed

$x=a\cos t,\ y=b\sin t$, so $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=a^2\sin^2t+b^2\cos^2t$. By symmetry, the perimeter is

\[P=4\int_0^{\pi/2}\sqrt{a^2\sin^2t+b^2\cos^2t}\,dt\]

Step 2: Rewrite the integrand

\[\sqrt{a^2\sin^2t+b^2\cos^2t}=b\sqrt{1+e^2\sin^2t},\qquad e^2=\frac{a^2-b^2}{b^2}\]

Numerically: $a^2\approx2.8284$, $b^2\approx2.4495$, so $e^2\approx0.1547$.

Step 3: Set up Simpson's rule on the quarter-integral

\[I=\int_0^{\pi/2}b\sqrt{1+e^2\sin^2t}\,dt,\qquad h=\frac{\pi/2}4=\frac\pi8\]

Evaluating $f(t)=b\sqrt{1+e^2\sin^2t}$ at $t=0,\frac\pi8,\frac\pi4,\frac{3\pi}8,\frac\pi2$:

\[f(0)\approx1.5651,\quad f\!\left(\tfrac\pi8\right)\approx1.5827,\quad f\!\left(\tfrac\pi4\right)\approx1.6245,\quad f\!\left(\tfrac{3\pi}8\right)\approx1.6652,\quad f\!\left(\tfrac\pi2\right)\approx1.6818\]

Step 4: Apply Simpson's rule

\[I\approx\frac h3\left[f_0+4f_1+2f_2+4f_3+f_4\right]\approx\frac{\pi/8}3[1.5651+6.3309+3.2490+6.6608+1.6818]\approx2.5509\]

Step 5: Total perimeter

\[P=4I\approx4(2.5509)\approx10.204\]

\[\boxed{P\approx10.20}\]

(This is close to the value obtained from direct high-precision numerical integration, $P\approx10.2037$, confirming Simpson's rule with just five points is already quite accurate here.)