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Question

Find the area enclosed by the parabolas y = 4x − x2 and y = x2 − x.

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Solution




We have, y=4x-x2 and y=x2-x
The points of intersection of two curves is obtained by solving the simultaneous equations

x2 -x=4x-x22x2 -5x =0 x=0 or x=52y=0 or y=154O0,0 and D 52 ,154 are points of intersection of two parabolas.In the shaded area CBDC , consider P(x,y2) on y =4x-x2 and Q(x,y1) on y=x2 -xAreaOBDCO= areaOBCO +areaCBDC=01y dx +152y2 -y1 dx=01y dx +152y2 -y1 dx y>0 y =y and y2 -y1y2-y1 as y2>y1 =014x-x2dx +1524x-x2-x2 -xdx=4x22-x3301+1525x-2x2dx=2x2 -x3301+5x22-2x33152=2-13+52522-23523 -52+23=53+5231-23 -116=53+52313-116=10-116+12524 =12124 sq units ...1AreaOCV'O =01y dx =01-y dx y<0 y =-y=01-x2 -xdx=01x-x2 dx=x22-x3 301=12-13=16 sq units ...2From 1 and 2Shaded area=areaOBDCOand areaOCV'O =12124+16 =12524 sq units

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