Browse Questions

# Evaluate the definite integral as limits of sums$\int\limits_a^bx\;dx$

$\begin{array}{1 1}\large \frac{b^2-a^2}{2} \\ \large \frac{b^2+a^2}{2} \\ \large \frac{a^2-b^2}{2} \\ \large \frac{b^2-a^2}{4} \end{array}$

Toolbox:
• $\int\limits_a^b f(x)dx=lim_{h->0}[f(a)+f(a+h)+...f(a+(n-1))]\;when\;h=\frac{b-a}{n}$
• $\int \limits_a^b f(x)dx=(b-a)\lim_{n \to \infty} \frac{1}{n}[f(a)+f(a+h)+f(a+(n-1)h)]$
• (ii)$\sum (n-1)=\frac{n(n-1)}{2}$
Given $I=\int \limits_a^b xdx$

we know $h=\frac{b-a}{n}$

Therefore \int \limits_a^b xdx =(b-a) \lim_{n \to \infty \frac{1}{n}[f(a)+f(a+h)+....+f(a+(n-1)h)]$After spliting a and h seperately we get$=(b-a) \lim _{n \to \infty} \frac{1}{n}[a+a+h+....+a+(n-1)h)]=(b-a) \lim _{n \to \infty} \frac{1}{n}[a+a+a+...nterm)+\frac{(b-a)}{n}(1+2+3+....(n-1)]$Since (a+a+a.....nterm)=na and$1+2+3+......(n-1)=\frac{n(n-1)}{2}$and$h=\frac{(b-a)}{n}$we can substitute n a and$\frac{n(n-1)}{2}$Therefore$(b-a) \lim _{n \to \infty} \frac{1}{n}[na+\frac{(b-a)}{n}\frac{(n)(n-1)}{2}]=(b-a) \lim _{n \to \infty} a +\frac{(n-1)}{2n}(b-a)]=(b-a) \lim _{n \to \infty} a +(\frac {1}{2}-\frac{1}{n})(b-a)]$applying the limits we get ,$(b-a)[a+\frac{1}{2}(b-a)]=\Large\frac{(b-a)(b+a)}{2}=\frac{b^2-a^2}{2}\$