# Let R be the region in the first and second quadrants bounded above by the graph of y=20/(1+x^2) and below by the horizontal line y=2, how do you find the area?

Jan 30, 2018

$\therefore R \approx 37.96$

#### Explanation:

Supposing $f \left(x\right) = \frac{20}{1 + {x}^{2}}$ and $g \left(x\right) = 2$, then to find $R$ is equivalent to finding ${\int}_{a}^{b} f \left(x\right) - g \left(x\right) \mathrm{dx}$ where $x = a$ and $x = b$ are the two $x$-values at which the functions $f$ and $g$ intersect.

This gives the area between the two curves (one curve and one line, rather). Function $f$ never goes below the $x$-axis, so the area is by definition restricted to the first two quadrants.

Start by integrating $f \left(x\right)$ using the trigonometric substitution $x = \tan \theta$ (recognising the Pythagorean Identity $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

$\int \frac{20}{1 + {x}^{2}} \mathrm{dx} \equiv 20 \cdot \int \frac{1}{\sec} ^ 2 \theta \mathrm{dx}$

Using the change of variable rule $\int f \left(x\right) \mathrm{dx} = f \left(x\right) \frac{\mathrm{dx}}{d \theta} d \theta$ we get $\frac{\mathrm{dx}}{d \theta} = {\sec}^{2} \theta$.

Hence $20 \cdot \int \frac{1}{\sec} ^ 2 \theta \mathrm{dx} \equiv 20 \cdot \int d \theta$

$\therefore \int f \left(x\right) \mathrm{dx} = 20 \arctan x + c$

$\int g \left(x\right) \mathrm{dx} = 2 x + c$, of course.

Now, we must find where $f$ and $g$ intersect by letting $f \left(x\right) = g \left(x\right)$.

$\frac{20}{1 + {x}^{2}} = 2$
$\therefore 20 = 2 \left(1 + {x}^{2}\right)$
$\therefore {x}^{2} - 9 = 0$
$\therefore x = 3$ or $- 3$.

Hence: $R = {\int}_{-} {3}^{3} f \left(x\right) \mathrm{dx} - {\int}_{-} {3}^{3} g \left(x\right) \mathrm{dx}$
$\therefore R = {\left[20 \arctan x\right]}_{-} {3}^{3} - {\left[2 x\right]}_{3}^{-} 3$
$= 20 \arctan 3 - 20 \arctan \left(- 3\right) - \left(6 + 6\right)$
$= 40 \arctan 3 - 12$ (exact answer).

$\therefore R \approx 37.96$