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Given :   and     Area of shaded region  = area of BCDB + are of ADCA                                                                                                                                                                                                                                        Hence, the correct answer is B.                                         
The area of the shaded region is to be found. Required area =ar(DOC)+ar(DOA) The region to the left of the y-axis is half of the circle with radius 4 units and centre origin. Area of the shaded region to the left of y axis is ar(1) =  For the region to the right of y-axis and above x axis       The parabola and the circle in the first quadrant intersect at point  Remaining area is 2ar(2)...
We have to find the area of the shaded region OCBAO Ar(OCBAO)=2ar(OCBO) For the fist quadrant In the first quadrant, the curves intersect at a point  Area of  the unshaded region in the first quadrant is The total area of the shaded region is- = Area of half circle - area of the shaded region in the first quadrant
We have to find the area of the shaded region ABC ar(ABC)=ar(ACLM)-ar(ALB)-ar(BMC) The lines intersect at points (1,2), (4,3) and (2,0) Area of the region bounded by the lines is 3.5 units
Equation of line joining A and B is Equation of line joining B and C is Equation of line joining A and C is ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM) ar(ABC)=8+5-6=7 Therefore the area of the triangle ABC is 7 units.
We have to find the area of the shaded region In the first quadrant y=|x|=x Area of the shaded region=2ar(OADO) The area bounded by the curves is 1/3 units.
We need to find the area of the shaded region ABCD ar(ABCD)=4ar(AOB) Coordinates of points A and B are (0,1) and (1,0) Equation of line through A and B is y=1-x The area bounded by the curve   is 2 units.
We have to find the area of the shaded region BAOB O is(0,0) The line and the parabola intersect in the second quadrant at (-1,1) The line y=x+2 intersects the x axis at (-2,0) The area of the region enclosed by the parabola  the line   and the -axis is 5/6 units.
We have to find the area of the shaded region The given ellipse and the given line intersect at following points Since the shaded region lies above x axis we take y to be positive The required area is
We have to find the area of the shaded region COB The two curves intersect at points (2,3) and (4,12) Required area is 
We have to find the area of the shaded region OBA The curves y=mx and y2=4ax intersect at the following points The required area is
The graph of y=sinx is as follows We need to find the area of the shaded region  ar(OAB)+ar(BCD) =2ar(OAB) The bounded area is 4 units.
y=|x+3| the given modulus function can be written as x+3>0 x>-3 for x>-3 y=|x+3|=x+3 x+3<0 x<-3 For x<-3 y=|x+3|=-(x+3) Integral to be evaluated is
the area of the region lying in the first quadrant and bounded by   and . The required area (ABCD) =                                            The area of the shaded region is 7/3 units
the area between the curves and . The curves intersect at A(1,1) Draw a normal to AC to OC(x-axis) therefore, the required area (OBAO)= area of (OCAO) - area of (OCABO)                                                            Thus the area of shaded region is 1/6 units
The area bounded  by the curev   and -axis The area of the required region = area of ABCD                                                     Hence the area of the shaded region is 624.8 units
The area bounded by the curve   and -axis The area of the required region = area of ABCD                                                     Hence the area of shaded region is 7/3 units
Simplifying the equation we get,                                 or                                                               Thus it becomes :                                                                                                                                                                                                                              
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