The area of the region (in sq. units) enclose | vedclass

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(A) The area $A$ is given by the integral $\int_{-1}^{1} | |x| - [x] | dx$. We split the integral into two parts: $\int_{-1}^{0} | |x| - [x] | dx + \i

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$2$

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(A) The area $A$ is given by the integral $\int_{-1}^{1} | |x| - [x] | dx$. We split the integral into two parts: $\int_{-1}^{0} | |x| - [x] | dx + \int_{0}^{1} | |x| - [x] | dx$. For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$. So,$| |x| - [x] | = | -x - (-1) | = | 1 - x | = 1 - x$. For $x \in [0, 1)$,$|x| = x$ and $[x] = 0$. So,$| |x| - [x] | = | x - 0 | = x$. At $x=1$,$|x|=1$ and $[x]=1$,so the difference is $0$. Thus,$A = \int_{-1}^{0} (1 - x) dx + \int_{0}^{1} x dx$. $A = [x - \frac{x^2}{2}]_{-1}^{0} + [\frac{x^2}{2}]_{0}^{1}$. $A = (0 - (-1 - \frac{1}{2})) + (\frac{1}{2} - 0) = (0 - (-\frac{3}{2})) + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = 2$ sq. units.

The area of the region (in sq. units) enclosed between the curves $y=|x|$,$y=[x]$ and the ordinates $x=-1$,$x=0$,$x=1$ is

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