Find the area enclosed by the curves y=p∣x∣−pandy=−1p∣x∣+py=p|x|-p\quad\text{and}\quad y=-\frac{1}{p}|x|+py=p∣x∣−pandy=−p1∣x∣+p when ppp is a positive constant. A4p3p2+1\dfrac{4p^3}{p^2+1}p2+14p3B2p3p2+1\dfrac{2p^3}{p^2+1}p2+12p3C8p3p2+1\dfrac{8p^3}{p^2+1}p2+18p3D4p3p−1\dfrac{4p^3}{p-1}p−14p3