find the area of the region bounded by the graphs of the given equations.\ny = 9x, y = x^3, x = 0, x =…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[0,3]$, we need to find which function is above the other. Let's find the difference $f(x)=9x - x^{3}$. When $x = 0$, $f(0)=0$; when $x = 3$, $f(3)=9\times3-3^{3}=27 - 27=0$. We can also take the derivative $f^\prime(x)=9 - 3x^{2}=3(3 - x^{2})$. The critical points are $x=\sqrt{3}$ and $x =-\sqrt{3}$. On the interval $[0,\sqrt{3}]$, $f^\prime(x)>0$ and on $[\sqrt{3},3]$, $f^\prime(x)<0$. For $x\in[0,\sqrt{3}]$, $y = 9x$ is above $y=x^{3}$, and for $x\in[\sqrt{3},3]$, $y=x^{3}$ is above $y = 9x$.

Step2: Set up the integral for the area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. So $A=\int_{0}^{\sqrt{3}}(9x - x^{3})dx+\int_{\sqrt{3}}^{3}(x^{3}-9x)dx$.

Step3: Integrate term - by - term

For $\int(9x - x^{3})dx=\frac{9x^{2}}{2}-\frac{x^{4}}{4}+C$. Evaluating $\int_{0}^{\sqrt{3}}(9x - x^{3})dx=\left[\frac{9x^{2}}{2}-\frac{x^{4}}{4}\right]{0}^{\sqrt{3}}=\frac{9\times3}{2}-\frac{9}{4}=\frac{27}{2}-\frac{9}{4}=\frac{54 - 9}{4}=\frac{45}{4}$. For $\int(x^{3}-9x)dx=\frac{x^{4}}{4}-\frac{9x^{2}}{2}+C$. Evaluating $\int{\sqrt{3}}^{3}(x^{3}-9x)dx=\left[\frac{x^{4}}{4}-\frac{9x^{2}}{2}\right]_{\sqrt{3}}^{3}=\left(\frac{81}{4}-\frac{81}{2}\right)-\left(\frac{9}{4}-\frac{27}{2}\right)=\frac{81 - 162}{4}-\frac{9 - 54}{4}=\frac{- 81}{4}-\frac{- 45}{4}=\frac{-81 + 45}{4}=-\frac{36}{4}$.

Step4: Calculate the total area

$A=\frac{45}{4}+\left|-\frac{36}{4}\right|=\frac{45 + 36}{4}=\frac{81}{4}$.

Answer:

$\frac{81}{4}$