The given equations are
y=x
and `y=x^3`
Using (i) in (ii) , we get
`x-x^3=0rArrx(1-x^2)=0rArr x(1-x)(1+x)=0`
`rArr x =0 or x =1 or x=-1`.
Also `(x=0 rArr y =0),(x= 1 rArr y=1)and (x=-1 rArr y =-1)`
So,the given curve and the line intersect at the points O(0,0)A(1,1) and B(-1,-1).
Now ,`y=x` is a line passing through the origin and making an angle of `45^@` with the x-axis .Thus the line y=x can be drawn.
For the curve `y=x^3` some values for x and the corresponding values of y are given below.
Plotiing the points (-1,-1)
`(-1/2,1/8),(0,0)(1/2,1/8), and
(1,1) and joining them , we get a rough sketch of `y=x^3` as shown in the given figure .
Required area
=(area ACOA)+(area ODBO)
=(area OALO)-(area OCALO)+(area OBMO)-(area ODBMO)
`=underset(0)overset(1)int "{y for (i)]dx"-underset(0)overset(1)int"{y for (ii)}dx" + underset(-1)overset(0){{(-y) for (i)}dx-underset(-1)overset(0)int{(-y)" for (ii) " }dx`
`underset(0)overset(1)int{(-y)" for (ii)}`dx
`underset(0)overset(1)int" x dx "-underset(0)overset(1)int x^3 dx+ underset(-1)overset(0)" -x xdx "- underset(-1)overset(0)-x^3 dx`
`=(1/2-1/4+1/2-1/4)=1/2` sq unit
Hence , the required area is 0.5 sq units.
