## Minimum Value of The Area Under The Curve

If y = f(x) is a strictly monotonic function in (a, b), with f'(x) ≠ 0, then the area of the region bounded by the ordinates x = a and x = b, function y = f(x) and y = f(c) (where c ∈ (a, b)) is minimum when c = (a + b)/2. (Consider the figure (fig. 1) given in the thumbnail).

Proof: Let us take y = f(x) strictly increasing function in the interval (a, b). The area of the shaded region shown in the figure (fig. 1) (in the thumbnail) is given by.

\begin{aligned}A=\int ^{c}_{a}\left( f\left( c\right) -f\left( x\right) \right) dx+\int ^{b}_{c}\left( f\left( x\right) -f\left( c\right) \right) dx\\ =f\left( c\right) \left( c-a\right) -\int ^{c}_{a}f\left( x\right) dx+\int ^{b}_{c}f\left( x\right) dx-f\left( c\right) \left( b-c\right) \\ =f\left( c\right) \left( 2c-\left( a+b\right) \right) +\int ^{b}_{c}f\left( x\right) dx-\int ^{c}_{a}f\left( x\right) dx\end{aligned}

Here area ‘A’ is the function of ‘c’. For maxima and minima f'(c) = 0. Now differentiating the equation given above with respect to ‘c’ thus we will have.

\begin{aligned}\dfrac{dA}{dc}=\left( 2c-\left( a+b\right) \right) f'\left( c\right) +2f\left( c\right) +0-2f\left( c\right) +0=0\\ \Rightarrow f'\left( c\right) \left( 2c-\left( a+b\right) \right) =0,f'\left( c\right) \neq 0\\ \Rightarrow 2c-\left( a+b\right) =0\Rightarrow c=\dfrac{a+b}{2}\\ \Rightarrow c <\dfrac{a+b}{2},\dfrac{dA}{dc} <0,c >\dfrac{a+b}{2},\dfrac{dA}{dc} >0\end{aligned}

Hence area ‘A’ is minimum when c = (a + b)/2. Now take a look at the following solved examples.

Example: If the area of the region bounded by the function f(x)= (x3)/3 – x2 + a, the x- axis and the straight lines x = 0, x = 2 and is minimum, then find the value of a.
Solution: Consider the figure (fig. 2 in the thumbnail) f'(x) = x2 – 2x = x (x – 2). Here f'(x) < 0 in (0, 2). Hence f(x) is strictly decreasing in (0, 2). So, for minimum area, f(x) must cross the x-axis at x = (0 + 2)/2 ⇒ x = 1. Hence f(1) = 1/3 - 1 + a = 0 ⇒ a = 2/3.

Example: Find the value of the parameter ‘a’ for which the area of the region bounded by the abscissa axis, the graph of the function y = x3 + 3x2 + 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, is the least.
Solution:
\begin{aligned}f\left( x\right) =x^{3}+3x^{2}+x+a\\ f'\left( x\right) =3x^{2}+6x+1\Rightarrow x=-1\pm \dfrac{\sqrt{6}}{3}\end{aligned}

Hence for the minimum area, the graph of the function must crosses x-axis at

\begin{aligned}x=\dfrac{1}{2}\left[ \left( -1+\dfrac{\sqrt{6}}{3}\right) +\left( -1-\dfrac{\sqrt{6}}{3}\right) \right] =-1\\ \Rightarrow f\left( -1\right) =-1+3-1+a=0\\ \Rightarrow a=-1\end{aligned}

Now attempt at least five questions based on this topic. So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Area Of The Region Bounded By Inequalities”.