If $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{cc}\sqrt{\left\{\mathrm{x}\right\}},& \mathrm{x}\notin \mathrm{Z}\\ 1,& \mathrm{x}\in \mathrm{Z}\end{array}\right.$and $g\left(x\right)=\{x{\}}^2$, (where {.} dentoes fractional part of x). Let R be the region bounded by f(x) and g(x) then

$f\left(x\right)=\left\{\begin{array}{cl}\sqrt{\left\{x\right\}},& x\notin z\\ 1,& x\in z\end{array}\text{ and }g\left(x\right)=\{x{\}}^2,\text{ where both }f(x)\right.\text{ and }g\left(x\right)\text{ are periodic with period ' }1\text{ ' shown as }$

a) Required area ${\int }_0^1 \sqrt{\mathrm{x}-\left[{\mathrm{x}}^{-}\right]}-{\int }_0^1 (\mathrm{x}-[\mathrm{x}]{)}^2\mathrm{dx}=1/3$

b) ${\int }_0^{10} \left[\mathrm{f}\right(\mathrm{x})-\mathrm{g}(\mathrm{x}\left)\right]\mathrm{dx}=10\left(\frac{1}{3}\right)=\frac{10}{3}$

(f(x) – g(x) in periodic function with period 1)

c) Required area = area of triangle – area R

$=5\left(1\right)-5\left(\frac{1}{3}\right)=\frac{10}{3}$

d) is wrong