1. ## Area under curve

Find the area under the curve y=05x−15

from x=6 to x=t and evaluate it for t=10 , t=100.
Then find the total area under this curve for x6.

The Integral is read as .5x^(-1.5)

(a) Find the area under the curve: .$\displaystyle y \:=\:\tfrac{1}{2}x^{-\frac{3}{2}}$ .from $\displaystyle x=6$ to $\displaystyle x=t.$
$\displaystyle A \;=\;\tfrac{1}{2}\int^t_6 x^{-\frac{3}{2}}dx \;=\;\tfrac{1}{2}\left(-\tfrac{2}{1}\right)x^{-\frac{1}{2}}\,\bigg]^t_6 \;=\;-\frac{1}{\sqrt{x}}\,\bigg]^t_6 \;=\;\left(-\frac{1}{\sqrt{t}}\right) - \left(-\frac{1}{\sqrt{6}}\right)$

Therefore: .$\displaystyle A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{t}}$

$\displaystyle \text{(b) Evaluate it for: }t=10,\: t=100$
$\displaystyle \text{For }t = 10\!:\;\;A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{10}}$

$\displaystyle \text{For }t = 100\!:\;\;A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{100}} \;=\;\frac{1}{\sqrt{6}} - \frac{1}{10}$

(c) Find the total area under this curve for $\displaystyle x \geq 6.$
This would be: .$\displaystyle \lim_{t\to\infty}\left(\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{t}}\right) \;\;=\;\;\frac{1}{\sqrt{6}} - 0 \;\;=\;\;\frac{1}{\sqrt{6}}$