Proof of the integrability of a step function

[stunner5000pt]

My second course in analysis and i have a problem which i can't understand

Let f (x) = 1 if 2<=x<4
2 if x =4
-3, if 4<x<=7
Prove that this function is integrable on [2,7], state its value and prove that it is what you say it is.
Obviously integral of f from [2,7] is -7. but its proof and the integrability have me and my friends snagged.

Suggestions anyone?

 

[HallsofIvy]

For any subdivision of [2,7] you can always choose a "finer" subdivision that includes 4 as a break point. That way you can isolate the discontinuity into two subdivisions, say [4-&delta;1,4] and [4,4+&delta;2]. What happens to the area of the rectangle based on those as &delta;1 and &delta;2 go to 0?

 

[M98Ranger]

Firstly, it is important to understand the definition and properties of integrability in order to prove that a function is integrable. A function is said to be integrable on a closed interval [a,b] if it is bounded and the set of its discontinuities has measure zero. In simpler terms, this means that the function must have a finite number of discontinuities and these points must not affect the overall area under the curve.

In the given function, f(x), we can see that it has three discontinuities at x=2, x=4, and x=7. However, these points do not affect the overall area under the curve as they are isolated points and do not form a continuous interval. Therefore, the set of discontinuities has measure zero and the function is bounded, making it integrable on [2,7].

To prove this, we can use the Riemann integral definition. We can divide the interval [2,7] into subintervals of equal length, and choose a partition P such that P = {x0, x1, x2, ..., xn} where x0=2, xn=7, and xi-xi-1 = h. Then, we can calculate the upper and lower sums for this partition as follows:

Upper sum, U(P,f) = ∑(xi-xi-1)sup(f(x)) = (2-2)sup(1) + (4-2)sup(2) + (7-4)sup(-3) = 0 + 4 + (-3) = 1
Lower sum, L(P,f) = ∑(xi-xi-1)inf(f(x)) = (2-2)inf(1) + (4-2)inf(2) + (7-4)inf(-3) = 0 + 2 + (-3) = -1

Now, as we take the limit of these sums as the partition gets finer and finer (i.e. h→0), we can see that both the upper and lower sums converge to the same value, which is -1. Therefore, we can conclude that the function is Riemann integrable on [2,7] and its integral is equal to -1.

In conclusion, the given function is integrable on [2,7] and its integral is equal to -1. This is because the function satisfies the