find the area of the region bounded by the graphs of the given equations. y = 6x², y = 6√x

Explanation:

Step1: Find intersection points

Set $6x^{2}=6\sqrt{x}$, then $x^{2}-\sqrt{x}=0$. Let $t = \sqrt{x}(t\geq0)$, so $t^{4}-t = 0$, $t(t^{3}-1)=0$. Solutions are $t = 0$ and $t = 1$, which means $x=0$ and $x = 1$.

Step2: Determine upper - lower functions

For $0<x<1$, $6\sqrt{x}\geq6x^{2}$.

Step3: Calculate the area

The area $A=\int_{0}^{1}(6\sqrt{x}-6x^{2})dx$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $A=\int_{0}^{1}(6x^{\frac{1}{2}}-6x^{2})dx=\left[6\times\frac{2}{3}x^{\frac{3}{2}}-6\times\frac{1}{3}x^{3}\right]{0}^{1}$. $A=(4x^{\frac{3}{2}} - 2x^{3})\big|{0}^{1}=4\times1^{\frac{3}{2}}-2\times1^{3}-(4\times0^{\frac{3}{2}}-2\times0^{3})=2$.

Answer:

$2$