Question

Evaluate the definite integral as limit of sums:
${\int }_0^5(x+1)dx$

Answer 1

${\int }_0^5(x+1)dx$

We know that

${\int }_a^{b}f\left(x\right)dx$

$=(b-a)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h\left)\right]$

Putting $a=0$ and $b=5$

$h=\frac{b-a}{n}=\frac{5-0}{n}=\frac{5}{n}$

$f\left(x\right)=x+1$

$\therefore {\int }_0^5(x+1)dx$

$=(5-0)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f(0+h)+f(0+2h)+...+f(0+(n-1)h\left)\right]$

$=5\underset{n\to \infty }{lim}\frac{1}{n}\left[f\right(0)+f(h)+f(2h)+...+f((n-1)h\left)\right]$

$=5\underset{n\to \infty }{lim}\frac{1}{n}(1+(h+1)+(2h+1)+..+(n-1)h+1)$

$=5\underset{n\to \infty }{lim}\frac{1}{n}(\underset{}{1+1+1+...+1}+h+2h+...+(n-1\left)h\right)$

$=5\underset{n\to \infty }{lim}\frac{1}{n}[n+h(1+2+...+(n-1)\left)\right]$

$=5\underset{n\to \infty }{lim}\frac{1}{n}(n+h\frac{(n-1)n}{2})$

$[\because 1+2+3+...+n=\frac{n(n+1)}{2}]$

$=5\underset{n\to \infty }{lim}(\frac{n}{n}+\frac{n(n-1)h}{2n})$

$=5\underset{n\to \infty }{lim}(1+\frac{n-1}{2}.\frac{5}{n})$

$(\text{Using}h=\frac{5}{n})$

$=5\underset{n\to \infty }{lim}[1+\frac{5}{2}(1-\frac{1}{n})]$

$=5[1+\frac{5}{2}(1-0\left)\right]$

$=5\left[\frac{7}{2}\right]$

$=\frac{35}{2}$