If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at `x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2)`.
`"Proof : " A=overset(c)underset(a)int(f(c)-f(x))dx+overset(b)underset(c)int(f(c))dx`
`=f(c)(c-a)-overset(c)underset(a)int(f(x))dx+overset(b)underset(a)int(f(x))dx-f(c)(b-c)`
`rArr" "A=[2c-(a+b)]f(c)+overset(b)underset(c)int(f(x))dx-overset(c)underset(a)int(f(x))dx`

Differentiating w.r.t. c, we get
`(dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0)`
For maxima and minima , `(dA)/(dc)=0`
`rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0)`
Hence, `c=(a+b)/(2)`
`"Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0`
Hence, A is minimum when `c=(a+b)/(2)`.
If the area bounded by `f(x)=(x^(3))/(3)-x^(2)+a` and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

Answer

`a=2/3`