Question

Find the area of the region bounded by the curve $y=x$ and the curve $y={{x}^{3}}$.

Answer 1

First draw the graph of the two functions and find the intersection points by solving the two equations. Once the point is determined, apply the formula for area bounded by two curves: $A=\int\limits_{a}^{b}{\left[ f(x)-g(x) \right]}dx$, where A is the area bounded between the curves, ‘a’ is the lower limit and ‘b’ is the upper limit. Here, ‘a’ and ‘b’ are the values of ‘x’, obtained after solving the equations. In the formula of area, f(x) is the graph of a function having higher value than the function g(x) in the range $\left[ a,b \right]$. When we will draw the graph of these functions, we will see that it is symmetric in the first and third quadrant. Therefore, there will be two similar regions of area, one in the 1st quadrant and the other in the 3rd quadrant. We have to determine the area of the region in the 1st quadrant and multiply it by 2 to get the total bounded area.

Complete step-by-step answer:

We have been provided with the two equations:

$y=x.................(i)$

$y={{x}^{3}}...............(ii)$

Let us solve these two equations to determine the point of intersection.

Equating the value of ‘y’ from equation (i) in equation (ii), we get,

$\begin{aligned}

& x={{x}^{3}} \\

& \Rightarrow {{x}^{3}}-x=0 \\

& \Rightarrow x\left( {{x}^{2}}-1 \right)=0 \\

\end{aligned}$

Substituting each term equal to 0, we get,

$\begin{aligned}

& x=0\text{ or }{{x}^{2}}=1 \\

& \Rightarrow x=0\text{ or }x=\pm 1 \\

\end{aligned}$

Hence, the lower limit is (x = -1) and the upper limit is (x = 1).

Let us draw the graph of the two functions.

As we can see that the graph is symmetric in the first and third quadrant, therefore, areas of the bounded regions in the two quadrants are equal. Therefore, we have to determine the area of the region in 1st quadrant and multiply it by 2.

Clearly we can see that the graph of the function $y=x$ is present above the graph of the function $y={{x}^{3}}$, in the range $\left[ 0,1 \right]$.

Therefore, $f(x)=x$ and $g(x)={{x}^{3}}$.

Now, using the formula for area of the curve bounded between the two functions: $A=\int\limits_{a}^{b}{\left[ f(x)-g(x) \right]}dx$, we get,

$A=2\int\limits_{0}^{1}{\left[ x-{{x}^{3}} \right]}dx$

Breaking the two terms, we get,

$\Rightarrow A=2\left[ \int\limits_{0}^{1}{x}dx-\int\limits_{0}^{1}{{{x}^{3}}}dx \right]$

Using the rule for integration given by: $\int{{{x}^{a}}dx}=\dfrac{{{x}^{a+1}}}{a+1}$, we get,

$\Rightarrow A=2\times \left[ \left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{{{x}^{4}}}{4} \right]_{0}^{1}$

Substituting the limits, we get,

& \Rightarrow A=2\times \left[ \left( \dfrac{{{1}^{2}}}{2} \right)-\dfrac{{{1}^{4}}}{4}-\left( \dfrac{{{0}^{4}}}{4}-\dfrac{{{0}^{4}}}{4} \right) \right] \\\ & \Rightarrow A=2\times \left[ \dfrac{1}{2}-\dfrac{1}{4} \right] \\\ & \Rightarrow A=2\times \left[ \dfrac{1}{4} \right] \\\ & \Rightarrow A=\dfrac{1}{2}\text{ sq}\text{. units} \\\ \end{aligned}$$ **Note:** One my note that, if we do not multiply the area of one region by 2 and take the area from the limit -1 to 1, we will get the same value because of the symmetric nature of the given functions. But this will be a lengthy process because then we will have to consider more conditions and therefore, the chances of making mistakes will be high.