Browse Questions

# Evaluate : $\int_0^2 (x^2+3)dx$ as limit of sum.

Toolbox:
• $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]$
• where $h=\large\frac{b-a}{n}$
• $\int \limits_a^b f(x)dx=\large\frac{(b-a)}{n}$$\lim_{n \to \infty} [f(a)+f(a+h)+f(a+(n-1)h)] • \sum (n-1)^2=\large\frac{n(n-1)(2n-1)}{6} \int \limits_a^b f(x)dx=\large\frac{(b-a)}{n}$$\lim_{n \to \infty} [f(a)+f(a+h)+f(a+(n-1)h)]$
where $h=\large\frac{b-a}{n}$
Here $a=0$ and $b=2$
$\therefore h=\large\frac{2}{n}$
$\therefore \int_0^2(x^2+3)dx=2\lim\limits_{n\to \infty}\large\frac{1}{n}$$[f(0)+f(\large\frac{2}{n})+f(\large\frac{4}{n}+........f(\large\frac{2(n-1)}{n})] \qquad\qquad\qquad=2\lim\limits_{n\to \infty}\large\frac{1}{n}$$[3+(\large\frac{2^2}{n^2+3})+(\large\frac{4^2}{n^2}+3)+........(\large\frac{(2n-2)^2}{n^2}+3)]$
$\qquad\qquad\qquad=2\lim\limits_{n\to \infty}\large\frac{1}{n}$$[(3+3+...3)+\large\frac{1}{n^2}$$(2^2+4^2+.........(2n-2)^2)]$
$\qquad\qquad\qquad=2\lim\limits_{n\to \infty}\large\frac{1}{n}$$[3n+\large\frac{2^2}{n^2}$$(1^2+2^2+.......(n-1)^2]$
$\qquad\qquad\qquad=2\lim\limits_{n\to \infty}\large\frac{1}{n}$$\bigg[3n+\large\frac{4}{n^2}\frac{n(n-1)(2n-1)}{6}\bigg] \qquad\qquad\qquad=2\lim\limits_{n\to \infty}\large\frac{1}{h}$$\bigg[3n+\large\frac{2}{3}\frac{(n-1)(2n-1)}{n}\bigg]$
$\qquad\qquad\qquad=2\lim\limits_{n\to \infty}[1+\large\frac{2}{3}(1-\large\frac{1}{h})(2-\large\frac{1}{h})]$
$\qquad\qquad\qquad=2[3+\large\frac{4}{3}]$
$\qquad\qquad\qquad=\large\frac{26}{3}$