Browse Questions

# Evaluate the definite integral as limits of sums$\int\limits_0^5(x+1)dx$

$\begin{array}{1 1}\large \frac{35}{2} \\35 \\\large \frac{45}{2} \\ 45\end{array}$

Toolbox:
• $\int\limits_a^b f(x)dx=lim_{h->0}h[f(a)+f(a+h)+...f(a+(n-1)h)]\;where\;h=\frac{b-a}{n}$
• $\int \limits_a^b f(x)dx=\frac{(b-a)}{n}\lim_{n \to \infty} [f(a)+f(a+h)+f(a+(n-1)h)]$
Given $I=\int \limits_0^5 (x+1)dx \qquad a=0\;and\;b=5$

we know $h=\frac{b-a}{n}=\frac{5-0}{n}\;and\;f(x)=x+1$

Therefore $h=\frac{5}{n}$

Therefore$\int \limits_0^5 (x+1)dx =\frac{5}{n} \lim_{n \to \infty} [f(a)+f(a+h)+....+f(a+(n-1)h)]$

$=\frac{5}{n} \lim_{n \to \infty} [f(0+1)+f(o+h)+....f(0+(n-1)h]$

$=\frac{5}{n}\lim_{n \to \infty}[(0+1)+(o+h+1)+(0+2h+1)....(0+(n-1)h+1]$

spliting the terms,

$=\frac{5}{n} \lim _{n \to \infty} [1+1+1+....nterm]+(h+2h+...(n-1)h]$

$=\frac{5}{n} \lim _{n \to \infty} [n+h(1+2+....(n-1)]$

substituting for $h=\frac{5}{h}\;and\;\sum(n-1)=\frac{n(n-1)}{2}$

$=\frac{5}{n} \lim _{n \to \infty}\{[n+\frac{5}{n}[\frac{n(n-1)}{2}]\}$

on simplifying,

$I=\frac{5}{n} \lim _{n \to \infty} [n+\frac{5}{2}(1-\frac{1}{n})]$

on applying limits,

$I=5+\frac{25}{2}$

$=\frac{35}{2}$