Question

Evaluate the definite integral as limit of sums: \(∫^5_0(x+1)dx\)

Answer 1

The correct answer is: \(\frac{35}{2}\)
Let \(I=∫^5_0(x+1)dx\)
It is known that,
\(∫^5_0ƒ(x)dx=(b-a)\underset{n→∞}{lim}\frac{1}{n}[ƒ(a)+ƒ(a+h)...ƒ(a+(n-1)h)]\),where \(h=\frac{b-a}{n}\)
Here,\(a=0,b=5\),and \(ƒ(x)=(x+1)\)
\(⇒h=\frac{5-0}{n}=\frac{5}{n}\)
\(∴∫^5_0(x+1)dx=(5-0)\underset{n→∞}{lim}\frac{1}{n}[ƒ(0)+ƒ(\frac{5}{n})+...+f((n-1)\frac{5}{n})]\)
\(=5\underset{n→∞}{lim}\frac{1}{n}[1+(\frac{5}{n}+1)+...[1+(\frac{5(n-1)}{n})]]\)
\(=5\underset{n→∞}{lim}\frac{1}{n}[(\underset{n times}{1+1+1...1})+[\frac{5}{n}+2.\frac{5}{n}+3.\frac{5}{n}+...(n-1)\frac{5}{n}]]\)
\(=5\underset{n→∞}{lim}\frac{1}{n}[n+\frac{5}{n}[1+2+3...(n-1)]]\)
\(=5\underset{n→∞}{lim}\frac{1}{n}[n+\frac{5}{n}.\frac{(n-1)n}{2}]\)
\(=5\underset{n→∞}{lim}\frac{1}{n}[n+.\frac{5(n-1)n}{2}]\)
\(=5\underset{n→∞}{lim}[1+\frac{5}{2}(1-\frac{1}{n})]\)
\(=5[1+\frac{5}{2}]\)
\(=5[\frac{7}{2}]\)
\(=\frac{35}{2}\)