First I need to find the area between the functions 2x^2-8x+11 and x^2-4x+10, then I need to find the area when it's rotated around the y-axis, assistance is much appreciated.
This one you should practice. Show your work. You do it by setting the two funcions equal solving setting up an integral based on the intersection coordinates and find within the interval of intersection which function is greater. Once you do that take the integral from the first intersection to the second of the greater function minus the smaller function'
show your work and we will critique
Ok so if you graphed your equations you would see they intersect at those points...and throughout that interval $\displaystyle (2-\sqrt{3},2+\sqrt{3})$ one is greater than the other
I instead of graphing will test point an element of the interval
$\displaystyle f(x)=2x^2-8x+11$
$\displaystyle g(x)=x^2-4x+10$
$\displaystyle f(0)=11$
$\displaystyle g(0)=10$
$\displaystyle \therefore,f(x)>g(x)\forall{x}\in(2-\sqrt{3},2+\sqrt{3})$
So your integral for the area would be
$\displaystyle \int_{2-\sqrt{3}}^{2+\sqrt{3}}f(x)-g(x)dx$