# What is the definite integral of zero?

Dec 17, 2014

If you mean ${\int}_{a}^{b} 0 \mathrm{dx}$, it is equal to zero.

This can be seen in a number of ways.

• Intuitively, the area under the graph of the null function is always zero, no matter over what interval we chose to evaluate it. Therefore, ${\int}_{a}^{b} 0 \mathrm{dx}$ should be equal to $0$, although this isn't an actual computation.

• Note the derivative of a constant function $\frac{d}{\mathrm{dx}} C = 0$.
By the Fundamental Theorem of Calculus, we get
${\int}_{a}^{b} 0 \mathrm{dx} = {\int}_{a}^{b} \frac{d}{\mathrm{dx}} C \mathrm{dx} = C \left(b\right) - C \left(a\right) = C - C = 0$

• Consider the Riemann Sums of the function $0$:
${\sum}_{i}^{n} f \left({x}_{i}\right) \Delta {x}_{i} = {\sum}_{i}^{n} 0 \Delta {x}_{i} ,$
where $\Delta {x}_{i}$ are the lengths of the divisions of the interval $\left[a , b\right]$.
No matter how we choose to divide the interval, this sum is always equal to $0$, since $0 \Delta {x}_{i} = 0$.
Therefore, the limit
${\lim}_{n \to \infty} {\sum}_{i}^{n} 0 \Delta {x}_{i} = {\int}_{a}^{b} 0 \mathrm{dx} = 0$