Question

help please 15..

Answer 1

@agent0smith x'd

Answer 2

http://www.wolframalpha.com/input/?i=y%3Dsec%5E2+x,+y%3D8+cos+x+from+x%3D-pi%2F3+to+x%3D+pi%2F3

Top minus bottom. \[\large \int\limits_{-\pi/3}^{\pi/3} (8 \cos x - \sec^2 x) dx\]or since it's symmetric \[\large 2\int\limits_{0}^{\pi/3} (8 \cos x - \sec^2 x) dx\]

Answer 3

Use these two facts
\[\Large \int \sec^2(x)dx = \tan(x)+C\]
\[\Large \int 8\cos(x)dx = 8\sin(x)+C\]
to help find the exact area

Answer 4

wait im lost D:

Answer 5

They're pretty nice that they're giving you endpoints, which happen to coincide exactly with where the equations intersect. How kind and un-owletgerish of them.

Answer 6

Marcie, it's just integral of top function minus bottom function. You don't even have to find intercepts.

Answer 7

What are you lost by? \[\large Area = \int\limits_{a}^{b} ( \text{\top function}- \text{bottom function} )dx\]

Answer 8

oh iight so then sec ^2x would be inserted as in the calculator as \[((\cos(x))^2\]

and yeh im confused by that top - bottom functions

Answer 9

That's how you find the area... you do the integral of the top function, minus the bottom function.

\[\large \sec^2 x = \frac{ 1 }{ (\cos x)^2 }\]
But you can prob do this w/o calculator.

Answer 10

oh okay thats where my mistake went

Answer 11

\[\large 2\int\limits\limits_{0}^{\pi/3} (8 \cos x - \sec^2 x) dx=\]
\[\Large 2\left[ 8 \sin x - \tan x\right]_0^{\pi/3}\]

Answer 12

oh okay i get it now x'd

Answer 13

Quit owletgin around on your calculator then and do it like a woman.

Answer 14

so how is it symmetric by multiplying it by 2 ?

Answer 15

lmfaoo x'd

Answer 16

Look at da graph http://www.wolframalpha.com/input/?i=y%3Dsec%5E2+x,+y%3D8+cos+x+from+x%3D-pi%2F3+to+x%3D+pi%2F3

Answer 17

I mean cos x is an even function and so is sec^2 x.. probably.

Answer 18

oh yeah :D

Answer 19

You don't have to use the symmetry, you can just do from -pi/3 to pi/3, it's just a little more thinking and effort.

Answer 20

iight thanks :D