# Show that

Question:

$\int_{0}^{5}(x+1) d x$

Solution:

$\operatorname{Let} I=\int_{0}^{5}(x+1) d x$

It is known that,

$\int_{a}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h) \ldots f(a+(n-1) h)]$, where $h=\frac{b-a}{n}$

Here, $a=0, b=5$, and $f(x)=(x+1)$

$\Rightarrow h=\frac{5-0}{n}=\frac{5}{n}$

$\therefore \int_{0}^{5}(x+1) d x=(5-0) \lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+\ldots+f\left((n-1) \frac{5}{n}\right)\right]$

$=5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[1+\left(\frac{5}{n}+1\right)+\ldots\left\{1+\left(\frac{5(n-1)}{n}\right)\right\}\right]$

$=5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[(1+1+1 \ldots 1)+\left[\frac{5}{n}+2 \cdot \frac{5}{n}+3 \cdot \frac{5}{n}+\ldots(n-1) \frac{5}{n}\right]\right]$

$=5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[n+\frac{5}{n}\{1+2+3 \ldots(n-1)\}\right]$

$=5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[n+\frac{5}{n} \cdot \frac{(n-1) n}{2}\right]$

$=5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[n+\frac{5(n-1)}{2}\right]$

$=5 \lim _{n \rightarrow \infty}\left[1+\frac{5}{2}\left(1-\frac{1}{n}\right)\right]$

$=5\left[1+\frac{5}{2}\right]$

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

$=\frac{35}{2}$