What is the definite integral of zero?

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Answer 1 · 2014-12-17T21:42:46

If you mean $\int_{a}^{b} 0 d x$ $int_a^b0dx$ , 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, $\int_{a}^{b} 0 d x$ $int_a^b 0 dx$ should be equal to $0$ $0$ , although this isn't an actual computation.
Note the derivative of a constant function $\frac{d}{d x} C = 0$ $d/(dx)C=0$ .
By the Fundamental Theorem of Calculus, we get
$\int_{a}^{b} 0 d x = \int_{a}^{b} \frac{d}{d x} C d x = C (b) - C (a) = C - C = 0$ $int_a^b 0 dx = int_a^b d/(dx) C dx = C(b) - C(a) = C - C = 0$
Consider the Riemann Sums of the function $0$ $0$ :
$sum_i^n f(x_i) Delta x_i = sum_i^n 0 Delta x_i ,$
where $Delta x_i$ are the lengths of the divisions of the interval $[a,b]$ .
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
$lim_(n to oo) sum_i^n 0 Delta x_i = int_a^b 0 dx = 0$