Proof of $\int_{a}^{a} f(x)dx = 0$

Question

For this proof, after I convert the definite integral into the riemann sum definition, is it just enough to say $\Delta(x)$ = $\frac{b-a}{n}$ and since b = a, the $\Delta(x)$ becomes 0, thus, making everything else equal to 0, since everything else is being multiplied by zero?

Answer 1

We can use also the empty sum definition $$\sum^0_{k=1} f(k)=0 $$ for every f.

One partition is in the form $a=t_0<t_1< \ldots <t_n=b$ if $a=b$ then $n=0$.

Let it be the degenerate interval $[a,a]=\{a\}$, we have $t_{0}=a$ and $t_{n}=a$, we should have $t_{1}>t_{0}=a$ but we don't have one number greater then $a$ in the interval. We have only one partition. The inferior summation is $$s(f,P)=\sum^{n=0}_{k=1}m_{k}\Delta t_{k-1}=0 $$ because it's empty. The superior summation is $$S(f,P)=\sum^{n=0}_{k=1}M_{k}\Delta t_{k-1}=0 $$ In this case the inferior summation is always equal the superior summation and the same for the superior and inferior integrals. So every function will be integrable in this set and the integral $0$ $$\int^{a}_{a} f(x)dx=0. $$

Answer 2

Why not just stick with this?

$$\begin{align} \int_a^a f(x)\, dx &= F(x)\bigr|_a^a \\ &= F(a)-F(a) \\ &= 0 \end{align}$$

where $F’(x)=f(x)$.