Question

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

Answer 1

The area is `9/4` square units.

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 `u = x^2`.

`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 `int_a^b` becomes `int_(1/2)^2`. Now draw a rudimentary graph of the two functions.

The graph in `color(blue)("blue")` is the parabola `y = 17/4 - x^2` and the graph in `color(red)("red")` is the rational function `y = x^(-2)`. As you can see, the parabola is above the rational function, therefore our expression that gives us the area will be

`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 `int_a^b F(x) dx = f(b) - f(a)`, where `F(x)` is continuous on `[a, b]` and `f'(x) = F(x)`.

`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!