Browse Questions

# Evaluate the following as limits of sums: $\int\limits_0^2(x^2+3)dx$

Toolbox:
• $\int \limits_a^b f(x)dx=h\; \lim_{ h \to 0} \bigg[f(a)+f(a+b)+....f(a)+(n-1)h \bigg]$
• where $h=\large\frac{b-a}{n}$
• $1^2+2^2+3^2+.....(n-1)^2=\large\frac{n(n-1)(2n-1)}{6}$
Let $I=\int \limits_0^2 (x^2+3)dx$
we have $\int \limits_a^b f(x)dx=h\; \lim_{ h \to 0} \bigg[f(a)+f(a+b)+f(a+2b)+....f(a+(n-1)h \bigg]$
Where $h=\large\frac{b-a}{n}$
Here $a=0\; and\; b=2\; and\; f(x)=x^2+3$
Therefore $h=\large\frac{2-0}{n}=\frac{2}{n}$
Therefore $\int \limits_0^2 (x^2+3) dx =\lim _{h \to 0} h \bigg[f(0)+f(0+h)+f(0+2h)+.........f(0+(n-1)h \bigg]$
$= \lim _{h \to 0} h \bigg[f(0)+f(h)+f(2h)+.........f((n-1)h \bigg]$
$=\lim _{h \to 0} h \bigg[(0+3)+(h^2+3)+(2^2h^2+3)+(3^2h^2+3).........(n-1)^2h^2+3 \bigg]$
But $3+3+3+.......... ntimes=3n$
$=\lim _{h \to 0} h \bigg[3n+h^2(1^2+2^2+.........(n-1)^2 \bigg]$
But $1^2+2^2+3^2+.....(n-1)^2=\large\frac{n(n-1)(2n-1)}{6}$
Substituting for h we get
$=\lim _{n \to \infty} \large\frac{2}{h} \bigg[3n+\frac{4}{h^2}.\frac{n(n-1)(2n-1)}{6} \bigg]$
$=\lim _{n \to \infty} \bigg[6+\large\frac{8}{6}.\frac{(n-1)(2n-1)}{n^2} \bigg]$
But $(n-1)=n(1-\frac{1}{h})\; and \;(2n-1)=n(2 -\frac{1}{n})$
$=\lim _{n \to \infty} \bigg[6+\large\frac{8}{6}.(1-\frac{1}{n})(2-\frac{1}{h}) \bigg]$
Now applying limits we get,
$=6+\large\frac{8}{6}(1-0)(2-0)=6+\frac{8}{3}$
$=\large\frac{26}{3}$