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

May 2, 2017

The area is $\frac{9}{4}$ square units.

#### Explanation:

Start by finding the points of intersection.

${x}^{- 2} = \frac{17}{4} - {x}^{2}$

$\frac{1}{x} ^ 2 = \frac{17}{4} - {x}^{2}$

$\frac{1}{x} ^ 2 + {x}^{2} - \frac{17}{4} = 0$

$4 {x}^{4} - 17 {x}^{2} + 4 = 0$

Now let $u = {x}^{2}$.

$4 {u}^{2} - 17 u + 4 = 0$

$4 {u}^{2} - 16 u - u + 4 = 0$

$4 u \left(u - 4\right) - \left(u - 4\right) = 0$

$\left(4 u - 1\right) \left(u - 4\right) = 0$

$u = \frac{1}{4} \mathmr{and} 4$

${x}^{2} = \frac{1}{4} \mathmr{and} {x}^{2} = 4$

$x = \pm \frac{1}{2} \mathmr{and} x = \pm 2$

Since we're talking about the area exclusively in the first quadrant, our standard integral ${\int}_{a}^{b}$ becomes ${\int}_{\frac{1}{2}}^{2}$. Now draw a rudimentary graph of the two functions.

The graph in $\textcolor{b l u e}{\text{blue}}$ is the parabola $y = \frac{17}{4} - {x}^{2}$ and the graph in $\textcolor{red}{\text{red}}$ 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

$A = {\int}_{\frac{1}{2}}^{2} \frac{17}{4} - {x}^{2} - \left({x}^{-} 2\right) \mathrm{dx}$

$A = {\left[\frac{17}{4} \left(x\right) - \frac{1}{3} {x}^{3} + {x}^{-} 1\right]}_{\frac{1}{2}}^{2}$

This can be evaluated using the second fundamental theorem of calculus which states that ${\int}_{a}^{b} F \left(x\right) \mathrm{dx} = f \left(b\right) - f \left(a\right)$, where $F \left(x\right)$ is continuous on $\left[a , b\right]$ and $f ' \left(x\right) = F \left(x\right)$.

$A = \frac{17}{4} \left(2\right) - \frac{1}{3} {\left(2\right)}^{3} + {2}^{-} 1 - \left(\frac{17}{4} \left(\frac{1}{2}\right) - \frac{1}{3} {\left(\frac{1}{2}\right)}^{3} + {\left(\frac{1}{2}\right)}^{-} 1\right)$

$A = \frac{17}{2} - \frac{8}{3} + \frac{1}{2} - \frac{17}{8} + \frac{1}{24} - 2$

$A = 7 - \frac{8}{3} - \frac{50}{24}$

$A = \frac{54}{24} = \frac{9}{4} \text{square units}$

Hopefully this helps!