Prove using riemann integral

Question

I dont really understand riemann integral. i have to prove that the integral of x^2 between 0 and 2 = 8/3

i understand that is equal to lim n->oo of 2/n summation (2i/n) but im unsure where to go from here?

Answer 1

You understood correctly. :-) In fact according to the definition of definite integral if $y=f(x)$ be a continuous function on interval $[a,b]$ then $$\int^a_bf(x)dx=\lim_{\Delta x\rightarrow0}\sum_{x=a}^bf(x)\Delta x$$ In a special numerical methods, based on dividing the interval into $n$ equal parts of lenght, we get $\Delta x=(b-a)/n$. So $$\int^a_bf(x)dx=\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum_{k=1}^nf\bigg( a+\frac{k(b-a)}{n}\bigg)$$ Now, we have $f(x)=x^2,~\frac{b-a}{n}=\frac{2}n$ and then $$\int_0^2x^2dx=\lim_{n\rightarrow\infty}\frac{2}{n}\sum_{k=1}^n\bigg(\frac{2k}{n}\bigg)^2=\lim_{n\rightarrow\infty}\frac{8}{n^3}\sum_{k=1}^nk^2=\lim_{n\rightarrow\infty}\left(\frac{8}{n^3}\times\frac{n(n+1)(2n+1)}{6}\right)=\frac{8}3$$