Area of Bounded Regions
4.0 Important Concept
4.0 Important Concept
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 $y=f(c)$ {$c \in (a,b)$}, is minimum when $c = \frac{{a + b}}{2}$
Question 9. If the area bounded by $f(x) = \frac{{{x^3}}}{3} - {x^2} + a$ and the straight lines $x=0$, $x=2$ and the $x$-axis is minimum, then find the value of '$a$'.
Solution: First to apply the above concept, the function should be a monotonic (increasing or decreasing) function.
We have $$\begin{equation} \begin{aligned} f(x) = \frac{{{x^3}}}{3} - {x^2} + a \\ \Rightarrow f'(x) = \frac{{3{x^2}}}{3} - 2x \\ \Rightarrow f'(x) = {x^2} - 2x \\ \Rightarrow f'(x) = x(x - 2) \\ \Rightarrow f'(x) < 0for(0,2) \\\end{aligned} \end{equation} $$Hence the function is monotonically decreasing in $(0,2)$.
The curve may be above or below $x$-axis or on $x$-axis. So, the shapes of curve can be like,
But for minimum possible area the curve must pass through $x$-axis. Therefore, the shape of the curve is
For the minimum area bounded $$\begin{equation} \begin{aligned} c = \frac{{0 + 2}}{2} \\ \Rightarrow c = 1 \\\end{aligned} \end{equation} $$
But, given that x-axis ($y=f(x)=0$) is that line with which the area is minimum. Hence,$$\begin{equation} \begin{aligned} y = f(1) = 0 \\ \Rightarrow \frac{{{{(1)}^3}}}{3} - {(1)^2} + a = 0 \\ \Rightarrow \frac{1}{3} - 1 + a = 0 \\ \Rightarrow a = \frac{2}{3} \\\end{aligned} \end{equation} $$
Question 10. Find the area of the region bounded by $\left| {\left| x \right| - \left| y \right|} \right| \leqslant 1$ and ${x^2} + {y^2} \leqslant 1$ in the $x-y$ coordinate plane.
Solution: From the given data, $$\begin{equation} \begin{aligned} \left| {\left| x \right| - \left| y \right|} \right| \leqslant 1 \\ \Rightarrow - 1 \leqslant \left| x \right| - \left| y \right| \leqslant 1 \\\end{aligned} \end{equation} $$ 
From the graph, it is clear that required area is area of circle.Therefore,$$\begin{equation} \begin{aligned} area(A) = \pi {r^2} \\ \Rightarrow A = \pi {(1)^2} \\ \Rightarrow A = \pi \\\end{aligned} \end{equation} $$
