Question

Let $f\left(x\right)=\{\begin{array}{c}\sqrt{\left\{x\right\}},x\notin Z\\ 1,x\in Z\end{array}$ and $g\left(x\right)=\{x{\}}^2,$ the area bounded by $f\left(x\right)$ and $g\left(x\right)$ for $x\in [0,n];(n\in Z)$ is denoted by $S\left(n\right)$, then which of the following is correct?

• A. $S\left(10\right)=\frac{5}{3}$
• B. $\frac{S\left(10\right)}{S\left(5\right)}=4$
• C. $\underset{n\to \infty }{lim}\frac{S\left(n\right)}{n}=\frac{1}{3}$
• D. $\sum _{n=1}^{9}S\left(n\right)=30$

Answer 1

The correct option is C $\underset{n\to \infty }{lim}\frac{S\left(n\right)}{n}=\frac{1}{3}$


The required area is,
$S\left(n\right)=n\underset{0}{\overset{1}{\int }}\sqrt{\left\{x\right\}}-\{x{\}}^{2}dx\Rightarrow S\left(n\right)=n\underset{0}{\overset{1}{\int }}\sqrt{x}-x^{2}dx\Rightarrow S\left(n\right)=n\times (\frac{2}{3}-\frac{1}{3})=\frac{n}{3}$
$S\left(10\right)=\frac{10}{3}\frac{S\left(10\right)}{S\left(5\right)}=\frac{10}{5}=2\underset{n\to \infty }{lim}\frac{S\left(n\right)}{n}=\underset{n\to \infty }{lim}\frac{n/3}{n}=\frac{1}{3}\sum _{n=1}^{9}S\left(n\right)=\frac{1}{3}\times \frac{9\times 10}{2}=15$