For completeness, we define an approximate integral that is also a linear spline, with a breakpoint sequence $z_{i+1}=\left(\nu_{i}+\nu_{i+1}\right) / 2$ for $i=2, \ldots, n$ and with additional break points defined by extrapolating beyond the original sequence: $z_{1}=\left(3 \nu_{1}-\nu_{2}\right) / 2$ and $z_{n+1}=\left(3 \nu_{n}-\nu_{n-1}\right) / 2$. The approximation to the integral,
$$F(x)=\int_{\nu_{1}}^{x} f(x) d x$$
at the new breakpoints is
$$F\left(z_{i}\right)=F\left(z_{i-1}\right)+\left(z_{i}-z_{i-1}\right) f\left(\nu_{i-1}\right),$$
where
$$F\left(z_{1}\right)=\frac{1}{2}\left(\nu_{1}-\nu_{2}\right) f\left(\nu_{1}\right)$$

(this ensures the normalization that $\left.F\left(\nu_{1}\right)=0\right) .{ }^{3}$ This definition produces an approximation to the integral at the original breakpoints that is equal to the approximation obt ained by applying the trapezoid rule :
$$\int_{\nu_{i}}^{\nu_{i+1}} f(x) d x \approx \frac{1}{2}\left(\nu_{i+1}-\nu_{i}\right)\left(f\left(\nu_{i+1}\right)+f\left(\nu_{i}\right)\right) .$$
(we leave the verification of this assertion as exercise for the reader).

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