find the area of the region bounded by the graphs of the given equations. y = 11x, y = x^2\nthe area is…

Explanation:

Step1: Find intersection points

Set $11x = x^{2}$, then $x^{2}-11x = 0$, factor out $x$ to get $x(x - 11)=0$. So $x = 0$ and $x = 11$ are the intersection - points.

Step2: Determine the upper and lower functions

For $0\leq x\leq11$, $y = 11x$ is above $y = x^{2}$.

Step3: Use the area formula

The area $A=\int_{a}^{b}[f(x)-g(x)]dx$, where $a = 0$, $b = 11$, $f(x)=11x$ and $g(x)=x^{2}$. So $A=\int_{0}^{11}(11x - x^{2})dx$.

Step4: Integrate

$\int(11x - x^{2})dx=\frac{11}{2}x^{2}-\frac{1}{3}x^{3}+C$. Then $A=\left[\frac{11}{2}x^{2}-\frac{1}{3}x^{3}\right]_{0}^{11}$.

Step5: Evaluate the definite - integral

$A=\frac{11}{2}\times11^{2}-\frac{1}{3}\times11^{3}=\frac{11^{3}}{2}-\frac{11^{3}}{3}=\frac{3\times11^{3}-2\times11^{3}}{6}=\frac{11^{3}}{6}=\frac{1331}{6}$.

Answer:

$\frac{1331}{6}$