Question

Find the area of the region $\left\{ \left( x,y \right):{{x}^{2}}\le y\le x \right\}$.

Answer 1

Hint: Observe that the region is bounded by two curves $y={{x}^{2}}$ and $y=x$. Identify the enclosed area by the two curves. Argue that the bounded area is equal to the difference between the area bounded by the curve $y=x$, the x-axis and the ordinates x =0 and x= 1 and the area bounded by the curve $y={{x}^{2}}$, the x-axis and the ordinates x= 0 and x= 1. Use the fact that the area bounded by the curve y = f(x) , the x-axis and the ordinates x = a and x= b is given by $y=\int_{a}^{b}{\left| f\left( x \right) \right|dx}$. Hence determine the two areas and hence the area of the region.

Complete step-by-step answer:
Let R be the region $\left\{ \left( x,y \right):{{x}^{2}}\le y\le x \right\}$
Hence in the region R, we have $y\ge {{x}^{2}}$ and $y\le x$
The region $y\ge {{x}^{2}}$ is shown below

The region $y\le x$ is shown below

Hence the region R is given by the intersection of these two regions

Hence the region R is the region ACBDA in the above diagram.
Finding the coordinates of A and B:
A and B are the points of intersection of the curves $y=x$ and $y={{x}^{2}}$
Hence, we have
${{x}^{2}}=x\Rightarrow x=0,1$
When x = 0, y = 0
Hence $A\equiv \left( 0,0 \right)$
When x = 1, y = 1
Hence $B\equiv \left( 1,1 \right)$
Observe that the area of the region R is the difference between the area bounded by the curve $y=x$, the x-axis and the ordinates x =0 and x= 1 and the area bounded by the curve $y={{x}^{2}}$, the x-axis and the ordinates x= 0 and x= 1.
Now, we know that the area bounded by the curve y = f(x) , the x-axis and the ordinates x = a and x= b is given by $y=\int_{a}^{b}{\left| f\left( x \right) \right|dx}$.
Hence the area bounded by the curve $y=x$, the x-axis and the ordinates x = 0 and x=1, is given by
${{A}_{1}}=\int_{0}^{1}{\left| x \right|dx}$
In the intervale (0,1), we have |x| = x
Hence, we have
${{A}_{1}}=\int_{0}^{1}{xdx}=\left. \dfrac{{{x}^{2}}}{2} \right|_{0}^{1}=\left( \dfrac{1}{2}-\dfrac{0}{2} \right)=\dfrac{1}{2}$
Also, the area bounded by the curve $y={{x}^{2}}$, the x-axis and the ordinates x = 0 and x =1, is given by
${{A}_{2}}=\int_{0}^{1}{\left| {{x}^{2}} \right|dx}$
We know that $\forall x\in R,\left| {{x}^{2}} \right|={{x}^{2}}$
Hence, we have
${{A}_{2}}=\int_{0}^{1}{{{x}^{2}}dx}=\left. \dfrac{{{x}^{3}}}{3} \right|_{0}^{1}=\left( \dfrac{1}{3}-\dfrac{0}{3} \right)=\dfrac{1}{3}$
Hence the area of the region R is given by $A={{A}_{1}}-{{A}_{2}}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$
Hence the area of the region R is $\dfrac{1}{6}$ square units.

Note: Alternative Solution:

Consider the vertical strip DECF
We have $DE=x-{{x}^{2}}$ and $CF=dx$
Hence the area of the strip is $\left( x-{{x}^{2}} \right)dx$
The total area of R is the sum of the areas of these strips from A to B
Hence, we have
$A=\int_{0}^{1}{\left( x-{{x}^{2}} \right)dx}=\left( \dfrac{1}{2}-\dfrac{1}{3} \right)=\dfrac{1}{6}$, which is the same as obtained above.