Question

How do you find the area of the region bounded by curves $y={{x}^{4}}$ and y=8x?

Answer 1

This type of problem is based on the concept of plotting graph and integration. First, we have to consider the given two equations. Equate both the equations and find the values of x. Plot a graph with the two equations and the intersection points. Observing from the graph, find the interval for which integration should be done for both the equations. On integrating, we get the area bounded which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to the area of the region bounded by curves $y={{x}^{4}}$ and y=8x.
We have been given the curves are $y={{x}^{4}}$ and y=8x.
On equating both the curves, we get
${{x}^{4}}=8x$
On adding -8x on both the sides of the equation, we get
${{x}^{4}}-8x=8x-8x$
On further simplification, we get
${{x}^{4}}-8x=0$
Here, we find that x is a common term. On taking x out as common, we get
$x\left( {{x}^{3}}-8 \right)=0$ ---------(1)
Since equation (1) is a product of two functions with variable x which is equal to 0, we get that the two functions are equal to 0.
Therefore, x=0 and ${{x}^{3}}-8=0$.
Now, let us consider${{x}^{3}}-8=0$. ---------(2)
Add 8 from both the sides of the equation (2). We get
${{x}^{3}}-8+8=0+8$
We know that the terms with the same magnitude and opposite sign cancel out.
Therefore, ${{x}^{3}}=8$.
Now, take the cube root on both the sides of the obtained equation. We get,
$\Rightarrow \sqrt[3]{{{x}^{3}}}=\sqrt[3]{8}$
We know that $\sqrt[3]{{{x}^{3}}}=x$ and $\sqrt[3]{8}=2$. Let us substitute this in the equation.
$\Rightarrow x=2$
Therefore, the values of x are 0 and 2.
Let us now plot a graph with the curves $y={{x}^{4}}$ and y=8x.

Here, the part shaded in yellow is the part bounded by two curves.
To find the area of the bounded part we have to integrate the curves with respect to x in the interval [0,2].
Since the curve y=8x is above the curve $y={{x}^{4}}$, we have to subtract the integration of $y={{x}^{4}}$ from y=8x in the interval [0,2] with respect to x.
Therefore, $Area=\int\limits_{0}^{2}{8xdx-}\int\limits_{0}^{2}{{{x}^{4}}dx}$ --------(3)
Let us first consider $\int\limits_{0}^{2}{8xdx}$.
$\int\limits_{0}^{2}{8xdx=}8\int\limits_{0}^{2}{xdx}$
Using the power rule of integration, that is, $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$. We get
$\int\limits_{0}^{2}{8xdx=}8\left[ \dfrac{{{x}^{1+1}}}{1+1} \right]_{0}^{2}$
On further simplification, we get
$\int\limits_{0}^{2}{8xdx=}8\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{2}$
On substituting the limits to the function, we get
$\int\limits_{0}^{2}{8xdx=}8\left[ \dfrac{{{2}^{2}}-0}{2} \right]$
$\Rightarrow \int\limits_{0}^{2}{8xdx=}8\left[ \dfrac{4}{2} \right]$
On cancelling out the common term from the numerator and denominator, we get
$\int\limits_{0}^{2}{8xdx=}4\times 4$
$\therefore \int\limits_{0}^{2}{8xdx=}16$
Now, let us consider $\int\limits_{0}^{2}{{{x}^{4}}dx}$.
Using the power rule of integration, that is, $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$. We get

$\int\limits_{0}^{2}{{{x}^{4}}dx}=\left[ \dfrac{{{x}^{4+1}}}{4+1} \right]_{0}^{2}$
On further simplification, we get
$\int\limits_{0}^{2}{{{x}^{4}}dx}=\left[ \dfrac{{{x}^{5}}}{5} \right]_{0}^{2}$
Substituting the limits to the integral, we get
$\int\limits_{0}^{2}{{{x}^{4}}dx}=\left[ \dfrac{{{2}^{5}}-0}{5} \right]$
We know that ${{2}^{5}}=32$.
Therefore, $\int\limits_{0}^{2}{{{x}^{4}}dx}=\left[ \dfrac{32}{5} \right]$.
Substituting both the values of integral in equation (3), we get
$Area=\int\limits_{0}^{2}{8xdx-}\int\limits_{0}^{2}{{{x}^{4}}dx}$
$\Rightarrow Area=16-\dfrac{32}{5}$
Taking LCM in the above expression, we get
$Area=\dfrac{16\times 5-32}{5}$
On further simplifications, we get
$\Rightarrow Area=\dfrac{80-32}{5}$
$\therefore Area=\dfrac{48}{5}$
Therefore, the area bounded by the curves $y={{x}^{4}}$ and y=8x is $\dfrac{48}{5}$ sq. units.

Note: We should not forget to put the units in the last step. Also avoid calculation mistakes based on sign conventions. Always the curve obtained in the graph should be integrated first and then the other curve should be integrated. We can also find the answer integrating the function with respect to y. For this method, we have to find the values of y from the given curves.