$$\lim_{N\to \infty} L_N = \lim_{N \to \infty} \Delta x \sum_{j=0}^{N-1} f(x_j) = \lim_{N \to \infty} \frac{4}{N} \sum_{j=0}^{N-1} (\frac {35}{2}[\frac{4j}{N} + 4])$$ $f(x) = (\frac{35}{2}(x+4))$ , $x_j = \frac{4j}{N}$ , $\Delta x = \frac{4}{N}$
I need to find the limit of $L_{N}$ at $[0,4]$ I've found $x_j$ and $\Delta x$ but am having trouble finding the limit.
On
A more basic way of handling this is to write the sum as $\frac{280}{N^2}\sum_{j= 0}^{N-1} j+ \frac{280}{N}\sum_{j= 0}^{N-1} 1$.
Adding 1 a total of N times (from 0 to N-1) is, of course, N so the second sum is just 280 which is independent of N. The first sum involves $\sum_{j=0}^{N-1} j= \sum_{j= 1}^{N-1} j$ (since the j= 0 term is just 0). It is well known, and easy to prove, that $\sum_{j= 1}^M j= \frac{M(M+1)}{2}$ so $\sum_{j= 0}^{N-1} j= \frac{(N-1)N}{2}= \frac{N^2- N}{2}$. $\frac{280}{N^2}\sum_{j=0}^{N-1} j= \frac{280(N^2- N)}{2N^2}= 140- \frac{140}{N}$. That obviously goes to 140 as N goes to infinity. Thus the initial limit is 280+ 140= 420.
Consider $f(x)=\frac{35}{2} (x+4)$, $f$ is continous on $[0,4]$, hence integrable there.
Divide $[0,4]$ into n equal part, i.e. $$0=\frac{4*0}{n}<\frac{4*1}{n}<...<\frac{4(n-1)}{n}<\frac{4*n}{n}=4$$ Note that each subinterval has width $$\Delta x=\frac{4}{n}$$ Pick $\xi_i=\frac{4}{n}$, we can form the following Riemann sum $$\sigma=\sum_{i=1}^{n}f(\xi_i)\Delta x$$ As $n \to \infty$, $\Delta x \to 0$, therefore $$\lim_{n \to \infty}\sum_{i=1}^{n}f(\xi_i)\Delta x=\int_{0}^{4}f(x)dx$$