Show that (1) the rectangle (midpoint) formula is exact

Show that (1) the rectangle (midpoint) formula is exact for $f(x)=ax+b$, and (2) the trapezoidal formula is exact for $f(x)=ax+b$.

Step 1: Exact integral

With $h=x_1-x_0$:

\[I_{\text{exact}}=\int_{x_0}^{x_1}(ax+b)\,dx=\frac{a}{2}(x_1^2-x_0^2)+b(x_1-x_0)=h\left[\frac{a}{2}(x_0+x_1)+b\right].\]

Step 2: Midpoint rule

The midpoint is $x_m=(x_0+x_1)/2$, so $f(x_m)=\dfrac{a}{2}(x_0+x_1)+b$, and

\[I_{\text{mid}}=h\cdot f(x_m)=h\left[\frac{a}{2}(x_0+x_1)+b\right]=I_{\text{exact}}.\]

\[\boxed{I_{\text{exact}}=I_{\text{mid}}}\]

Step 3: Trapezoidal rule

\[I_{\text{trap}}=\frac{h}{2}\left[f(x_0)+f(x_1)\right]=\frac{h}{2}\left[a(x_0+x_1)+2b\right]=h\left[\frac{a}{2}(x_0+x_1)+b\right]=I_{\text{exact}}.\]

\[\boxed{I_{\text{exact}}=I_{\text{trap}}}\]

Step 4: Why both work

For a linear function, the midpoint value equals the average value of $f$ over the interval, and so does the average of the two endpoint values — both facts are special to linear (degree-1) functions, which is exactly why both rules recover the true area exactly, geometrically the area under a straight line is itself a trapezoid, matched exactly by either construction.

Step 5: Numerical check

For $f(x)=3x+2$ on $[1,5]$: $I_{\text{exact}}=44$; midpoint with $x_m=3$ gives $4\times f(3)=4\times11=44$; trapezoidal gives $2\times[f(1)+f(5)]=2\times[5+17]=44$. Both match exactly.