Find The Area Bounded by the given curves. How would I set this up to solve?

x=2, x=5, y=1/x, y=1/x^2

Here is what your graph looks like

http://www.wolframalpha.com/input/?i=plot+y%3D1%2Fx,+y%3D1%2Fx%5E2

look at the second diagram in quadrant 1

Between x = 2 and x = 5, which is the higher curve?
So you want
∫ (higher function - lower function) dx from x = 2 to x = 5

A soccer ball is kicked from the ground in an arc defined by the function,h(x) = -2x2 + 8x. What is the height of the ball after 3 seconds?

To find the area bounded by the given curves, you can set up an integral that represents the difference between the two curves. Here's how you can set it up:

First, sketch the curves on a coordinate plane to get a visual representation of what you're trying to find.

The given curves are x = 2, x = 5, y = 1/x, and y = 1/x^2.

The curve y = 1/x is a hyperbola that starts from the positive y-axis and extends to the positive x-axis. The curve y = 1/x^2 is another hyperbola that lies entirely in the first quadrant.

To find the area bounded by the curves, you need to find the points where the curves intersect. In this case, the curves intersect at (2, 1/2), (2, 1/4), and (5, 1/5) as shown on the sketch.

Next, you need to determine the limits of integration. Since the curves intersect at x = 2 and x = 5, these will be your lower and upper limits, respectively.

To find the area between the curves, you need to subtract the area under the curve y = 1/x^2 from the area under the curve y = 1/x. The integral setup would be as follows:

Area = ∫ [1/x - 1/x^2] dx

To evaluate this integral, you can simplify the expression using algebraic manipulations. The resulting integral will depend on the limits of integration (2 to 5).

Finally, compute the definite integral using antiderivatives or by using the appropriate numerical methods like numerical integration or approximation techniques.

Once you have set up the integral and simplified it, you can proceed to evaluate the integral to find the area bounded by the given curves.