Proof of Various Integral Properties — Question 2

Question 2

Assume that $f$ and $g$ are integrable on the interval $[a,b]$. Prove that $$\int_a^b \bigl(f(x)+g(x)\bigr)\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx.$$

Question 2 - Solution

We prove this property using Riemann sums.

Let $$a=x_0<x_1<\cdots<x_n=b$$ be a partition of $[a,b]$, and let $\Delta x_i=x_i-x_{i-1}$. Choose a sample point $x_i^*$ in each subinterval $[x_{i-1},x_i]$.

A Riemann sum for $f+g$ is $$\sum_{i=1}^n \bigl(f(x_i^*)+g(x_i^*)\bigr)\Delta x_i.$$

Using algebra, split the sum: $$\sum_{i=1}^n \bigl(f(x_i^*)+g(x_i^*)\bigr)\Delta x_i = \sum_{i=1}^n f(x_i^*)\Delta x_i + \sum_{i=1}^n g(x_i^*)\Delta x_i.$$

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

Since $f$ and $g$ are integrable on $[a,b]$, the limits of both sums exist: $$\lim_{\|P\|\to 0}\sum_{i=1}^n f(x_i^*)\Delta x_i = \int_a^b f(x)\,dx,$$ $$\lim_{\|P\|\to 0}\sum_{i=1}^n g(x_i^*)\Delta x_i = \int_a^b g(x)\,dx.$$

Therefore, $$\int_a^b \bigl(f(x)+g(x)\bigr)\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx.$$

$$\boxed{\int_a^b (f+g)=\int_a^b f+\int_a^b g}$$