Question

If f(x) is monotonic in (a,b) then prove that the area bounded by the ordinates at $x=a:x=b:y=f\left(x\right)$ and $y=f\left(c\right),cϵ(a,b)$ is minimum when $c=\frac{a+b}{2}.$
Hence if the area bounded by the graph of $f\left(x\right)=\frac{x^3}{3}-x^2+a,$ the straight lines $x=0$, $x=2$ and the x-axis is minimum then find the value or 'a'.

• A. $\frac{2}{3}$
• B. $\frac{2}{5}$
• C. $\frac{7}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{2}{3}$

Let us consider the function $f\left(x\right)$ which is monotonically increasing in $(a,b)$

Consider a point $c$ between $aandb$

$a<c<b$

$f\left(a\right)<f\left(c\right)<f\left(b\right)$ (f(c) is constant)

$Areaofshadedregion={\int }_a^{c}f\left(c\right)dx-{\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx-{\int }_c^{b}f\left(c\right)dx$

$A=f\left(c\right){\int }_a^{c}dx-f\left(c\right){\int }_c^{b}dx+{\int }_c^{b}f\left(x\right)dx-{\int }_a^{c}f\left(x\right)dx$

$A=(c-a)f\left(c\right)-(b-c)f\left(c\right)+{\int }_c^{b}f\left(x\right)dx+{\int }_c^{a}f\left(x\right)dx$

$A=[2c-(a+b\left)\right]f\left(c\right)+{\int }_c^{b}f\left(x\right)dx+{\int }_c^{a}f\left(x\right)dx$

We know that ${\int }_c^{b}f\left(x\right)dx+{\int }_c^{a}f\left(x\right)dx$ is positive

Area will be minimum when [2c-(a+b)]f(c)=0

$f\left(c\right)\ne 0$

2c-(a+b)=0

$c=\frac{(a+b)}{2}$

$A={\int }_0^{2}f\left(x\right)dx={\int }_0^2[\frac{x^3}{3}-x^2+a]dx={\int }_0^2\frac{x^3}{3}dx-{\int }_0^2{x}^{2}dx+{\int }_0^{2}adx$

$A={\left[\frac{x^4}{12}\right]}_0^2-{\left[\frac{x^3}{3}\right]}_0^2+a{\left[x\right]}_0^2$

$A=\frac{12}{3}-\frac{8}{3}+2a$

=$(2a-\frac{4}{3})$

Minimum area $2a-\frac{4}{3}$=0

$a=\frac{2}{3}$