Krizalid said:This one is pretty tricky, so write $n-k$ as a sum and reverse the order of the sums.
[sp]Let \(\displaystyle F(x) = \int_0^xf(t)\,dt.\) Then $$\frac{1}{n}\sum_{k = 0}^{n - 1} (n - k)\int_{k/n}^{(k+1)/n}\!\!\! f(x)\,dx = \frac1n \sum_{k = 0}^{n - 1} (n-k)\bigl(F(\tfrac{k+1}n\bigr) - F\bigl(\tfrac kn\bigr)\bigr) = \frac1n \sum_{k = 0}^{n - 1} (n-k)F(\tfrac{k+1}n\bigr) - \frac1n \sum_{k = 0}^{n - 1}(n-k) F\bigl(\tfrac kn\bigr).$$ The first of those two sums is $$\sum_{j = 0}^{n - 1} (n-j)F(\tfrac{j+1}n\bigr) = \sum_{k=1}^{n} (n-k+1)F(\tfrac{k}n\bigr) = F(1) - F(0) + \sum_{k=0}^{n-1} (n-k+1)F(\tfrac{k}n\bigr)$$ (first writing $j$ instead of $k$, and then letting $k=j+1$). Therefore $$\begin{aligned}\frac{1}{n}\sum_{k = 0}^{n - 1} (n - k)\int_{k/n}^{(k+1)/n}\!\!\! f(x)\,dx &= \frac1n\Bigl(F(1) - F(0) + \sum_{k=0}^{n-1} (n-k+1)F(\tfrac{k}n\bigr)\Bigr) - \frac1n \sum_{k = 0}^{n - 1}(n-k) F\bigl(\tfrac kn\bigr) \\ &= \frac{F(1)}n + \frac1n\sum_{k=0}^{n-1}F(\tfrac{k}n\bigr).\end{aligned}$$ As $n\to\infty$, $F(1)/n\to0$ and the Riemann sum \(\displaystyle \frac1n\sum_{k=0}^{n-1}F(\tfrac{k}n\bigr)\) converges to \(\displaystyle \int_0^1F(x)\,dx.\) But (integrating by parts) \(\displaystyle \int_0^1 {(1 - x)f(x)\,dx} = \Bigl[(1-x)F(x)\Bigr]_0^1 + \int_0^1F(x)\,dx = \int_0^1F(x)\,dx.\) Put those results together to see that $$\lim_{n\to\infty}\frac1n \sum_{k=0}^n (n-k) \int_{k/n}^{(k+1)/n}\!\!\!f(x)\,dx = \int_0^1(1-x)f(x)\,dx.$$[/sp]Krizalid said:For any integrable function on $[0,1]$ prove that $\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 0}^{n - 1} {(n - k)\int_{\frac{k}{n}}^{\frac{{k + 1}}{n}} {f(x)\,dx} } = \int_0^1 {(1 - x)f(x)\,dx} .$
An integrable function on $[0,1]$ is a function that can be integrated over the interval from 0 to 1. This means that the area under the curve of the function over this interval is finite.
The limit of an integrable function on $[0,1]$ is defined as the value that the function approaches as the independent variable approaches a specific value within the given interval. In other words, it is the value that the function "converges" to as the input approaches a certain point.
Proving a limit for an integrable function on $[0,1]$ is important because it helps us understand the behavior of the function near a specific point within the given interval. It also allows us to determine if the function is continuous at that point, which is a key concept in calculus.
Some common methods used to prove a limit for an integrable function on $[0,1]$ include the epsilon-delta method, the squeeze theorem, and using properties of limits such as algebraic manipulation and the sandwich theorem.
Yes, a limit of an integrable function on $[0,1]$ can exist even if the function is not continuous at that point. This is because continuity is a necessary but not sufficient condition for a limit to exist. However, if a function is not continuous at a point, the limit at that point will not be equal to the value of the function at that point.