Question

Find the area of the region enclosed by the parabola x 2 = y , the line y = x + 2 and x -axis

Answer 1

We have to find the area enclosed by the parabola whose equation is x 2 =y , the line y=x+2 and x-axis. Draw the graphs of the equations and shade the common region.

Figure (1)

Solve the equation of parabola and straight line to find the points of intersection.

y=x+2

Substitute this value of y in equation of parabola.

x 2 =x+2 x 2 −x−2=0

Further, solve the above equation.

x 2 −2x+x−2=0 x( x−2 )+( x−2 )=0 ( x+1 )( x−2 )=0 x=−1,2

Corresponding values of y are,

y= ( −1 ) 2 , ( 2 ) 2 y=1,4

The coordinates of point A are ( −1,1 ) .

Since the point of intersection is known, drop a perpendicular from the points of intersection to the x-axis.

The area of the region OBAO is,

Area of the region OBAO=Area of the region OCAO+Area of the region CBAC

To find the area bound by the straight line y=x+2 with the x-axis, assume a vertical strip of infinitesimally small width and integrate the area.

Area of the region OCAO= ∫ −2 −1 ( x+2 ) dx = [ x 2 2 +2x ] −2 −1 =[ ( −1 ) 2 2 +2( −1 )−( ( −2 ) 2 2 +2( −2 ) ) ] = 1 2

Similarly find the area bound by the parabola with x-axis,

Area of the region CBAC= ∫ −1 0 y dx

From the equation of parabola, find the value of y in terms of x and substitute in the above integral.

Area of the region CAOBDC= ∫ −1 0 x 2 dx = [ x 3 3 ] −1 0 = 1 3 [ ( 0 ) 3 − ( −1 ) 3 ] = 1 3

Total area of the shaded region is,

Area of the region OBAO=Area of the region OCAO+Area of the region CBAC = 1 2 + 1 3 = 5 6

Thus, the area enclosed by the parabola whose equation is x 2 =y , and the line y=x+2 is 5 6 sq units .