Is f(x) integrable on the interval -1 < x < 1?

[Swamifez]

Let

f(x)=

{1 if -1 < x<0;
{-1 if 0 < x < 1.

Prove that f(x) is integrable on -1 < x < 1.

 

[quasar987]

Just show that the superior and inferior integrals are equal.

 

[tacman]

To prove that f(x) is integrable on the interval -1 < x < 1, we must show that the upper and lower Riemann sums approach the same value as the partition size approaches 0.

Let P = {x0, x1, ..., xn} be a partition of the interval -1 < x < 1, where xi = -1 + iΔx and Δx = (1 - (-1))/n.

The upper Riemann sum is defined as U(f,P) = Σf(xi*)Δx, where xi* is any point in the subinterval [xi-1, xi]. Similarly, the lower Riemann sum is defined as L(f,P) = Σf(xi**)Δx, where xi** is any point in the subinterval [xi-1, xi].

For any partition P, we have:
U(f,P) - L(f,P) = Σ[f(xi*) - f(xi**)]Δx

Now, let's consider the subintervals [xi-1, xi] for i = 1, 2, ..., n. In each of these subintervals, we have either f(xi*) = 1 and f(xi**) = -1, or f(xi*) = -1 and f(xi**) = 1. This means that for each subinterval, f(xi*) - f(xi**) = 2 or -2.

Therefore, we can rewrite U(f,P) - L(f,P) as:
U(f,P) - L(f,P) = 2ΣΔx = 2(1-(-1)) = 4

As n approaches infinity, Δx approaches 0, and hence U(f,P) and L(f,P) approach the same value of 2. This shows that f(x) is integrable on the interval -1 < x < 1, since the upper and lower Riemann sums approach the same value as the partition size approaches 0.

In conclusion, f(x) is indeed integrable on the interval -1 < x < 1.