Question

help please

sketch the region enclosed by the given curves and find its area

Answer 1

\[y= \tan x\]\[y= 2sinx \]\[-\pi/3\le x \le \pi/3\]

Answer 2

sketch: http://www.wolframalpha.com/input/?i=y%3Dtan+x,+y%3D2sin+x+from+x%3D-pi%2F3+to+x%3Dpi%2F3
Just split it into two integrals, then do top function minus bottom function, for each portion of the area.\[\large \int\limits_{-\pi/3}^{0} (\tan x - 2 \sin x)dx+\int\limits_{0}^{\pi/3} (2 \sin x - \tan x)dx\]
You could also just find one of those, and double it. It looks (and should be) pretty symmetric from the graphs of sine and tangent.

Answer 3

iight so all i know is that ill have to set it equal to each other lol

\[tanx = 2\sin x \]

Answer 4

The graph saves you time in even doing that. You can see the intersections must be /pi/3, 0 and pi/3.

Answer 5

oh i see it x'd

Answer 6

how come the graph looks different from my calculator ?

Answer 7

idk, maybe you got it in degree mode, or maybe your calculator sucks

Answer 8

haha. its in radian mode D:

Answer 9

Snapchat that graph

Answer 10

hmm wait i just zoomed it in .. i guess it looks good right ??

Answer 11

Your y-axis sucks wingspan, use the same scale as in the link i gave

Answer 12

true x'd so what would my y axis be lool zzz

Answer 13

The same as what they use lol x'd

Answer 14

x'd waittt i think i got it x'd

Answer 15

aah i hate these graphs x'd

Answer 16

lol why you hate em

Answer 17

they too much problems lmaoo

Answer 18

loool x'd too owletgerish

Answer 19

owleterishhh as flutter

Answer 20

lol you understand dis one tho??

Answer 21

x'D

Answer 22

x'd

Answer 23

yeh i understand it neegaerishhh x'd

Answer 24

x'd

Answer 25

aight neegarishhh x'd