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=int_(a)^(c) (f(c)-f(x))dx+int_(c)^(b) (f(c))dx`
`=f(c)(c-a)-int_(a)^(c) (f(x))dx+int_(a)^(b) (f(x))dx-f(c)(b-c)`
`rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b) (f(x))dx-int_(a)^(c) (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)`.
The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function `y=x^(3)+3x^(2)+x+a`, and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is