Proof of Various Integral Properties — Question 8

Question 8

Assume that $f$ is integrable on $[a,b]$ and that $c$ satisfies $$a<c<b.$$ Prove that $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.$$

Question 8 - Solution

We prove this property using Riemann sums.

Let $$a=x_0<x_1<\cdots<x_k=c<\cdots<x_n=b$$ be a partition of $[a,b]$ that includes the point $c$.

Let $$\Delta x_i = x_i-x_{i-1}$$ and choose a sample point $x_i^*$ in each subinterval $[x_{i-1},x_i]$.

A Riemann sum for $f$ over $[a,b]$ is $$\sum_{i=1}^n f(x_i^*)\,\Delta x_i.$$

Split this sum at $c$: $$\sum_{i=1}^n f(x_i^*)\,\Delta x_i = \sum_{i=1}^k f(x_i^*)\,\Delta x_i + \sum_{i=k+1}^n f(x_i^*)\,\Delta x_i.$$

The first sum is a Riemann sum for $f$ over $[a,c]$, and the second sum is a Riemann sum for $f$ over $[c,b]$.

Now take the limit as the norm of the partition $\|P\|\to 0$.

Since $f$ is integrable on $[a,b]$, it is integrable on both subintervals $[a,c]$ and $[c,b]$.

Thus, $$\lim_{\|P\|\to 0}\sum_{i=1}^k f(x_i^*)\,\Delta x_i = \int_a^c f(x)\,dx,$$ $$\lim_{\|P\|\to 0}\sum_{i=k+1}^n f(x_i^*)\,\Delta x_i = \int_c^b f(x)\,dx.$$

Therefore, $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.$$

$$\boxed{\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx}$$