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 $ƒ(x)=(x+1)$
$⇒h=\frac{5-0}{n}=\frac{5}{n}$
$∴∫^5_0(x+1)dx=(5-0)\underset{n→∞}{lim}\frac{1}{n}[ƒ(0)+ƒ(\frac{5}{n})+...+f((n-1)\frac{5}{n})]$
$=5\underset{n→∞}{lim}\frac{1}{n}[1+(\frac{5}{n}+1)+...[1+(\frac{5(n-1)}{n})]]$
$=5\underset{n→∞}{lim}\frac{1}{n}[(\underset{n times}{1+1+1...1})+[\frac{5}{n}+2.\frac{5}{n}+3.\frac{5}{n}+...(n-1)\frac{5}{n}]]$
$=5\underset{n→∞}{lim}\frac{1}{n}[n+\frac{5}{n}[1+2+3...(n-1)]]$
$=5\underset{n→∞}{lim}\frac{1}{n}[n+\frac{5}{n}.\frac{(n-1)n}{2}]$
$=5\underset{n→∞}{lim}\frac{1}{n}[n+.\frac{5(n-1)n}{2}]$
$=5\underset{n→∞}{lim}[1+\frac{5}{2}(1-\frac{1}{n})]$
$=5[1+\frac{5}{2}]$
$=5[\frac{7}{2}]$
$=\frac{35}{2}$