Answer:
23.92
Step-by-step explanation:
We solve for y:
[tex]25x^2-4y^2=100[/tex]
[tex]y=5\sqrt{x^4/4-1}[/tex]
use trig substitution:
[tex]x=2secu[/tex]
[tex]x^2=4sec^2u[/tex]
The derivative of x is:
[tex]dx=2secutanu du[/tex]
when x=2 [tex]u=0[/tex]
when x=3 [tex]u=sec^-^1\cdot(3/2)[/tex]
The area is defined as 2xarea:
The area is the integral of the equation:
[tex]A=2\int\limits^a_b {5\sqrt{x^2/4-1} } \, dx[/tex] for range 0 to sec-1(3/2)
Substitute x=2secu
[tex]A=\int\limits^a_b 10{\sqrt{(4/4)sec^2u-1} } \, dx[/tex]
[tex]A=10\int\limits^a_b {\sqrt{sec^2i-1}23secutanu } \, du[/tex]
We know that sec²-1 = tan²u
[tex]A=20\int\limits^a_b {tan^2u secu} \, du[/tex]
[tex]A=20\int\limits^a_b {(sec^2u-1)secu} \, du[/tex]
[tex]A=20[\int\limits^a_b {sec^3u} \, du - \int\limits^a_b {secu} \, du][/tex]
After simplifying
[tex]A=10[secutanu-ln(secu+tanu)][/tex]
For the range
[tex]A=10[(3/2)\sqrt{5} /2-ln(3/2+\sqrt{5}/2)][/tex]
[tex]A=23.92[/tex]
