Answer
$\int_{0}^{6} (-x^2+6x) dx$
Work Step by Step
Find the solutions to $x^2-6x=0$, which are $x=[0,6]$. Therefore, the integral must go from 0 to 6. Next, find the function that is greater over the interval, which is $y_{2}=0$. Therefore, use the definition of area between curves, or $\int_{a}^{b} (g(x)-f(x)) dx$, where $g(x)$ is greater, to reach a final answer of $\int_{0}^{6} (0-(x^2-6x)) dx$ = $\int_{0}^{6} (-x^2+6x) dx$.
