**P164. Show that the area of the bounded region enclosed between the curves \((y^3=x^2)\) and \((y=2-x^2)\), is \((2\frac{2}{15})\).**

**Solution:**

Note that \((y=x^{\frac{2}{3}})\) is an even function (green line).

**P165. Find the area of the region in the xy plane, bounded by the graphs of \((y=x^2)\), x+y = 2 and \((y=-\sqrt {x})\)**

**Solution:**

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

\((\int^1_0 x^2,dx=\left[\frac{x^3}{3}\right]^1_0=\frac{1}{3})\) (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 = \((\frac{1}{2}\times 1\times 1=\frac{1}{2})\)

Thus total area above x axis (of the required region) is \((\frac{1}{2}+\frac{1}{3}=\frac{5}{6})\)

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

x+y = 2 and \((y=-\sqrt{x})\) intersect at (4, -2) (found by solving the two equations). We calculate the area under the curve \((y=-\sqrt{x})\) 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

\((|\int^4_0 -\sqrt{x},dx|=\int^4_0 \sqrt{x},dx= \left[\frac {x^{\frac{1}{2}+1}} {\frac{1}{2}+1}\right]^4_0)\).

=\((\frac {2}{3}\times 8=\frac {16}{3})\)

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

**Area = \((\frac {16}{3} – 2 + \frac{5}{6} = \frac{25}{6})\) (ANS)**

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