Approximate pi/4=integral from 0 to 1 dx/(1+x²) using (a)

Approximate $\dfrac{\pi}{4}=\displaystyle\int_0^1\frac{dx}{1+x^2}$ using (a) the midpoint rule with $n=10$, (b) the trapezoidal rule with $n=10$, and (c) Simpson's rule with $n=2$; estimate the error in each case.

Step 0: Exact value

\[I_{\text{exact}}=\int_0^1\frac{dx}{1+x^2}=\big[\arctan x\big]_0^1=\frac{\pi}{4}\approx0.785398163.\]

Step 1(a): Midpoint rule, $n=10$

With $h=0.1$ and midpoints $x_i=(i-\tfrac12)h$:

Summing the ten values gives $\sum f(x_i)\approx7.856065$, so

\[I_{\text{mid}}=h\sum f(x_i)\approx0.1\times7.856065=0.785606.\]

\[\boxed{I_{\text{mid}}\approx0.785606}\]

Step 1(b): Error bound for the midpoint rule

The midpoint-rule error bound is $E_{\text{mid}}\le\dfrac{(b-a)^3}{24n^2}\max|f''(x)|$. With $f''(x)=\dfrac{6x^2-2}{(1+x^2)^3}$, one finds $|f''(x)|$ is largest at $x=0$, where $|f''(0)|=2$. So

\[E_{\text{mid}}\le\frac{1}{24\times100}\times2=\frac{1}{1200}\approx0.000833.\]

The actual error is $|0.785606-0.785398|\approx0.000208$, comfortably inside the bound.

Step 2(a): Trapezoidal rule, $n=10$

With nodes $x_i=0,0.1,\dots,1.0$ and $f(x_i)$ computed at each, the trapezoidal sum is

\[I_{\text{trap}}=\frac{h}{2}\left[f(0)+f(1)+2\sum_{i=1}^{9}f(x_i)\right]=\frac{0.1}{2}\times15.699630\approx0.784981.\]

\[\boxed{I_{\text{trap}}\approx0.784981}\]

Step 2(b): Error bound for the trapezoidal rule

\[E_{\text{trap}}\le\frac{(b-a)^3}{12n^2}\max|f''(x)|=\frac{1}{1200}\times2=\frac{1}{600}\approx0.001667.\]

Actual error: $|0.784981-0.785398|\approx0.000417$, again inside the bound.

Step 3(a): Simpson's rule, $n=2$

With $h=0.5$, nodes $0,0.5,1$: $f(0)=1$, $f(0.5)=0.8$, $f(1)=0.5$.

\[I_{\text{Simpson}}=\frac{0.5}{3}\left[1+4(0.8)+0.5\right]=\frac{0.5}{3}\times4.7=0.783333.\]

\[\boxed{I_{\text{Simpson}}\approx0.783333}\]

Step 3(b): Error bound for Simpson's rule

\[E_{\text{Simpson}}\le\frac{(b-a)^5}{180n^4}\max|f^{(4)}(x)|.\]

One computes $f^{(4)}(x)=\dfrac{24(5x^4-10x^2+1)}{(1+x^2)^5}$, which attains its maximum magnitude $24$ at $x=0$. With $n=2$:

\[E_{\text{Simpson}}\le\frac{1}{180\times16}\times24=\frac{1}{120}\approx0.008333.\]

Actual error: $|0.783333-0.785398|\approx0.002065$, inside the bound.

Step 4: Summary comparison

All three approximations bracket $\pi/4$ within their theoretical error bounds; with these particular choices of $n$, the midpoint rule (using ten subintervals) happens to be the most accurate, followed by the trapezoidal rule, with Simpson's rule using only two subintervals the least accurate in absolute terms — a reminder that a higher-order method with very few points is not automatically better than a lower-order method with many points.