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Q.
Area bounded by the lines $y = |x| - 2$ and $y = 1 - |x - 1|$ is equal to
Application of Integrals
Solution:
We have, $y = - x - 2 \quad ...(i) $
$y = x - 2 \quad ....(ii)$
$ y = 2 - x \quad .....(iii)$
$ y = x \quad ... (iv)$
Solving $(iii)$ and $(iv)$, we get $A(1, 1)$
Solving $(i)$ and $(iv)$ we get $D(-1, -1)$
Required area = area of $\Delta AOB$ + area of $\Delta OCB $
+ area of $\Delta OCD$
Area of $\Delta AOB=\int\limits_{0}^{1} x \,dx +\int\limits_{1}^{2} \left(2-x\right) dx$
$ = \left[\frac{x^{2}}{2}\right]_{0}^{1} +\left[2x-\frac{x^{2}}{2}\right]_{1}^{2} $
$= \frac{1}{2} + \left[4-\frac{4}{2}\right] -\left[2-\frac{1}{2}\right]$
$ = \frac{1}{2} +\frac{4}{2} -\frac{3}{2} = 1$ sq. units
Area of $\Delta OCB = \left|\int\limits_{0}^{2} \left(x-2\right) dx \right|$
$ = \left|\left[\frac{x^{2}}{2} -2x\right]_{0}^{2}\right| = 2$ sq. units
Area of $ \Delta OCD = \left|\int\limits_{-1}^{0}\left(-x-2\right) dx -\int\limits_{-1}^{0}x\, dx \right| $
$=\left|-\left[\frac{x^{2}}{2} + 2x\right]_{-1}^{0} -\left[\frac{x^{2}}{2}\right]_{-1}^{0}\right| = 1$ sq. unit
