# How do you find the area in the first quadrant bounded by #y=x^-2# and #y=17/4 - x^2#?

##### 1 Answer

The area is

#### Explanation:

Start by finding the points of intersection.

#x^(-2) = 17/4 - x^2#

#1/x^2 = 17/4 - x^2#

#1/x^2 + x^2 - 17/4= 0#

#4x^4 - 17x^2 + 4 = 0#

Now let

#4u^2 - 17u + 4 = 0#

#4u^2 - 16u - u + 4 = 0#

#4u(u - 4) - (u - 4) = 0#

#(4u - 1)(u - 4) = 0#

#u = 1/4 or 4#

#x^2 = 1/4 or x^2 = 4#

#x = +- 1/2 or x = +-2#

Since we're talking about the area exclusively in the first quadrant, our standard integral

The graph in

#A = int_(1/2)^2 17/4 - x^2 - (x^-2) dx#

#A = [17/4(x) - 1/3x^3 + x^-1]_(1/2)^2#

This can be evaluated using the second fundamental theorem of calculus which states that

#A = 17/4(2) - 1/3(2)^3 + 2^-1 - (17/4(1/2) - 1/3(1/2)^3 + (1/2)^-1)#

#A = 17/2 - 8/3 + 1/2 - 17/8 + 1/24 - 2#

#A = 7 - 8/3 - 50/24#

#A = 54/24 = 9/4 "square units"#

Hopefully this helps!