Solveeit Logo
Login

Question

Mathematics Question on integral

Evaluate the definite integral as limit of sums: ∫23x2dx∫^3_2 x^2dx

Answer

The correct answer is: 193\frac{19}{3}
It is known that,
∫abƒ(x)dx=(b−a)limn→∞1n[ƒ(a)+ƒ(a+h)+ƒ(a+2h)....ƒa+(n−1)h],∫^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}],whereh=b−anh=\frac{b-a}{n}
Here,a=2,b=3,a=2,b=3,and ƒ(x)=x2ƒ(x)=x^2
⇒h=3−2n=1n⇒h=\frac{3-2}{n}=\frac{1}{n}
∴∫23x2dx=(3−2)limn→∞1n[ƒ(2)+ƒ(2+1n)+ƒ(2+2n)...ƒ[2+(n−1)1n]]∴∫^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}]]
=limn→∞1n[(2)2+(2+1n)2+(2+2n)2+...(2+(n−1)2n)]=\underset{n→∞}{lim}\frac{1}{n}[(2)^2+(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+...(2+\frac{(n-1)^2}{n})]
=limn→∞1n[22+[22+(1n)2+2.2.1n]+...+[(2)2+(n−1)2n2+2.2.(n−1)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}]]
=limn→∞1n[(22+....+22ntimes)+(1n)2+(2n)2+...+(n−1n)2+2.2.[1n+2n+3n+...+(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}]]
=limn→∞1n[4n+1n2[12+22+32+...+(n−1)2]+4n[1+2+...+(n−1)]]=\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)]]
=limn→∞1n[4n+1n2[n(n−1)(2n−1)6]+4n[n(n−1)2]=\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}]
=limn→∞1n[[4n+n(1−1n)(2−1n)6+4n−42]=\underset{n→∞}{lim}\frac{1}{n}[[4n+\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6}+\frac{4n-4}{2}]
=limn→∞[4+16(1−1n)(2−1n)+2−2n]=\underset{n→∞}{lim}[4+\frac{1}{6}(1-\frac{1}{n})(2-\frac{1}{n})+2-\frac{2}{n}]
=4+26+2=4+\frac{2}{6}+2
=193=\frac{19}{3}