Consider the bounded region R between the graphs of $f(x) = x^2$ and $g(x) = 3x$. Sketch R, which appropriate labels. Then set up all the integrals required do not evaluate.
My solution:
Definitions:
$A = \displaystyle\int_{a}^{b} (f(x) - g(x))dx$
$M_{x} = y-\text{bar} = \frac{1}{A}\displaystyle\int_{a}^{b} \frac{1}{2}(f(x)^{2} - g(x)^{2})dx$
$M_{y} = x-\text{bar} = \frac{1}{A}\displaystyle\int_{a}^{b} x(f(x) - g(x))dx$
….
$A = \displaystyle\int_{0}^{3} (x^{2} - 3x)dx = \frac{9}{2}$
$M_{x} = y-\text{bar} = \frac{2}{9}\displaystyle\int_{0}^{3} \frac{1}{2}((x^{2})^{2} - (3x)^{2})dx$
$M_{y} = x-\text{bar} = \frac{2}{9}\displaystyle\int_{0}^{3} x(x^{2} - 3x)dx$
The graph was straightforward. Am not sure if these are right or wrong. Please help Thank you
The correct integrals should be $$A=\int_0^3(3x-x^2)\,dx=\frac92$$ $$M_x=\frac29\int_0^3x(3x-x^2)\,dx$$ $$M_y=\frac29\int_0^3\frac{(3x)^2-(x^2)^2}2\,dx$$ since $3x>x^2$ over $(0,3)$.