**P164. Show that the area of the bounded region enclosed between the curves and , is .**

**Solution:**

Note that is an even function (green line).

**P165. Find the area of the region in the xy plane, bounded by the graphs of , x+y = 2 and **

**Solution:**

The parabola and straight line intersects at (1,1) (we find that by solving the and x+y=2)

Thus the area is found by adding area under parabola (from 0 to 1) and area under straight line (from 1 to 2).

(area under parabola)

area under straight line above ‘x’ axis is the triangle with height 1 unit and base 1 unit (from x=1 to x=2, area under x+y=2)

that area =

Thus total area above x axis (of the required region) is

Now we come to the region below ‘x’ axis.

x+y = 2 and intersect at (4, -2) (found by solving the two equations). We calculate the area under the curve from x=0 to x=4 and subtract from it the area of the triangle with base from x=2 to x=4 and height =2 (hence the area of the triangle to be subtracted is 2 sq unit).

Area under the square root curve is

.

=

Delete 2 square unit from this and add the area computed before (above ‘x’ axis).

**Area = (ANS)**