# Area between $x^3$ and $\sqrt{x}$

I'm having some difficult to find the intersection between $$x^3$$ and $$\sqrt{x}$$ for calculate the area between them. Could someone help me?

• For both $x=0$ and $x=1$, we have $x^3 = x^{1/3}$. – mjw Jul 1 at 4:34
• For one of the roots of $x^{1/3}$, we also have $x=-1$, which will be a point of intersection on the graph $x$ vs $f(x)$ with $x$ and $f(x)$ real. – mjw Jul 1 at 4:34
• But how do you get this result? – Mycroft Jul 1 at 4:35
• How to get what result: $\{-1,0,1\}$? – mjw Jul 1 at 4:36
• This is not how I solved it (I just saw that $x^3$ and $x^{1/3}$ are inverses so they are symmetric with respect to the line $x=y$. Anyway, cube both sides: $x^9-x=0$. You can factor this. $x(x^8-1)=0$. This has roots $x=0$ and the "eigth roots of unity" of which two are real: $x\in\{-1,1\}$. – mjw Jul 1 at 4:40

## 3 Answers

$$\begin{equation} x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0 \end{equation}$$ And so we have $$x = 0$$ or $$x^8 - 1 = 0$$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus, $$\begin{equation} x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0 \end{equation}$$ Assuming you are seeking Real Solutions only we see that $$x^4 + 1 = 0$$ has no solutions. For $$x^4 - 1$$ employ the same identity: $$\begin{equation} (x^4 + 1)(x^4 - 1) = 0 \rightarrow (x^4 + 1)(x^2 + 1)(x^2 - 1) = 0 \end{equation}$$ As with $$x^4 + 1 = 0$$ having no Real Solutions we also observe that $$x^2 + 1 = 0$$ also have no Real Solutions. Thus the only remaining Real Solutions as those that satisfy: $$\begin{equation} x^2 - 1 = 0 \rightarrow x = \pm 1 \end{equation}$$ As such, the three intersection points of $$x^3$$ and $$x^{\frac{1}{3}}$$ occur at $$x = -1, 0, 1$$

To find the intersection between the graph, just find the solutions of $$x^3=x^{1/3}$$. $$x^3=x^{1/3} \Rightarrow x^9-x=x(x-1)(x+1)(x^2+1)(x^4+1)=0 \Rightarrow x=-1,0,1$$ Since $$x^3\geq x^{1/3}$$ for $$x\in[-1,0]$$, and $$x^{1/3}\geq x^3$$ for $$x\in[0,1]$$, the area between the two graphs is given by $$\int_{-1}^0 (x^3-x^{1/3})dx+\int_0^1 (x^{1/3}-x^3)dx$$

Solve the equation $$x^3=\sqrt{x}$$:

$$x^3=x^{\frac13}\implies\\ \left(x^3\right)^3=\left(x^{\frac13}\right)^3\implies\\ x^9=x.$$

Divide both sides by $$x$$ and observe that $$x=0$$ is a solution:

$$\frac{x^9}{x}=1\implies\\ x^{9-1}=1\implies\\ x^8=1\implies\\ x=\pm1.$$

So, the solution set consists of three elements: $$\{-1,0,1\}$$. And those are the $$x$$ values for your intersection points.

• Nice. I like it. – mjw Jul 1 at 4:46