Question

Evaluate the definite integral as limit of sums: \(∫^3_2 x^2dx\)

Answer 1

The correct answer is: \(\frac{19}{3}\)
It is known that,
\(∫^b_a ƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)+ƒ(a+2h)....ƒ{a+(n-1)h}],\)where\(h=\frac{b-a}{n}\)
Here,\(a=2,b=3,\)and \(ƒ(x)=x^2\)
\(⇒h=\frac{3-2}{n}=\frac{1}{n}\)
\(∴∫^3_2 x^2dx=(3-2)\underset{n→∞}{lim}\frac{1}{n}[ƒ(2)+ƒ(2+\frac{1}{n})+ƒ(2+\frac{2}{n})...ƒ[2+(n-1)\frac{1}{n}]]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[(2)^2+(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+...(2+\frac{(n-1)^2}{n})]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[2^2+[2^2+(\frac{1}{n})^2+2.2.\frac{1}{n}]+...+[(2)^2+\frac{(n-1)^2}{n^2}+2.2.\frac{(n-1)}{n}]]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[(\underset{n times}{2^2+....+2^2})+{(\frac{1}{n})^2+(\frac{2}{n})^2+...+(\frac{n-1}{n})^2}+2.2.[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{(n-1)}{n}]]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[4n+\frac{1}{n^2}[1^2+2^2+3^2+...+(n-1)^2]+\frac{4}{n}[1+2+...+(n-1)]]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[4n+\frac{1}{n^2}[\frac{n(n-1)(2n-1)}{6}]+\frac{4}{n}[\frac{n(n-1)}{2}]\)
\(=\underset{n→∞}{lim}\frac{1}{n}[[4n+\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}+\frac{4n-4}{2}]\)
\(=\underset{n→∞}{lim}[4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}]\)
\(=4+\frac{2}{6}+2\)
\(=\frac{19}{3}\)