Proof of Various Integral Properties — Question 4

Question 4

Assume that $f$ is integrable on $[a,b]$. Prove that $$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx.$$

Question 4 - Solution

We use the definition of the definite integral in terms of Riemann sums.

Let $$a=x_0<x_1<\cdots<x_n=b$$ be a partition of the interval $[a,b]$, with subinterval widths $$\Delta x_i=x_i-x_{i-1}.$$

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

Now consider the interval $[b,a]$. Reverse the order of the partition: $$b=x_n>x_{n-1}>\cdots>x_0=a.$$

The corresponding subinterval widths are $$\Delta x_i' = x_{i-1}-x_i = -\Delta x_i.$$

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

Take the limit as the norm of the partition goes to zero. Since $f$ is integrable on $[a,b]$, the limit exists: $$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx.$$

Rewriting, $$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx.$$

$$\boxed{\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx}$$