Sorry, need help again.. The question is:
Caulculate the area bounded by the graphs of f(x) = x^2, g(x) = 1/x^2, x>0 and the line x = 3.
First equate the two equations to see where the functions intersect.
$\displaystyle x^2 = \frac{1}{x^2}$
which is x= 1
That i going to be the lower bound.
Now notice how x^2 is above 1/x^2 in the interval [1,3]
so subtract the lower function from the higher function
put all that together and you get
$\displaystyle \int_1^3 x^2 - \frac{1}{x^2}$