Question

Find area between the curves. Please show me the steps! I got confused somewhere in the middle.

Answer 1

\[\int\limits_{0}^{\pi/6}(2\cos2(x)-2\sin(x))dx+\int\limits_{\pi/6}^{\pi/2}(2\sin(x)-2\cos2(x))dx\] is how I set this up o.o

Answer 2

sec -_- eating... nom nom nom

Answer 3

no worries xD take your time. Food comes first.

Answer 4

were those the only two graphs provided?

Answer 5

em no there are two more heh

Answer 6

But i'm pretty sure it's the one on the right in the screenshot.

Answer 7

Sorry slow today :p

Hmmm your set up looks good.

Answer 8

https://www.wolframalpha.com/input/?i=integral+from+x%3D0+to+x%3Dpi%2F2+of+abs%282sinx-2cos%282x%29%29dx

https://www.wolframalpha.com/input/?i=%28integral+from+x%3D0+to+x%3Dpi%2F6+of+2cos%282x%29-2sinx+dx%29%2B%28integral+from+x%3Dpi%2F6+to+x%3Dpi%2F2+of+2sinx-2cos%282x%29+dx%29

Answer 9

Ok so wolfram at least verifies for us that your set up is correct :) cool

Answer 10

okii and then I integrated and got:
\[[\sin2(x)+2\cos(x)] (0 \rightarrow \pi/6)+[2\cos(x)-\sin2(x)](\pi/6 \rightarrow \pi/2)\]
and of course those aren't multiplied by the (0 -> pi/6)
I just don't know how to do the evaluated at thingy xD

Answer 11

Ya it's weird :)
You can do |_{0}^{pi/6} \(|_{0}^{\pi/6}\)
_ is for subscript
^ is for exponent

Answer 12

ooooooo\[|_{0}^{\pi/6}\]
feeling professional x) thanks!! haha

Answer 13

Ok I see boo boo I think.

Answer 14

2sinx --> -2cosx when integrating, ya?

Answer 15

Yeah

Answer 16

Ok So I guess we should've had this:\[[\sin2(x)+2\cos(x)] (0 \rightarrow \pi/6)+[\color{orangered}{-2\cos(x)}-\sin2(x)](\pi/6 \rightarrow \pi/2)\]You use WAY too many brackets by the way lol

Answer 17

Like, I would've written it like this...\[\large\rm \left[\sin2x+2\cos x\right]_0^{\pi/6}+\left[-2\cos x-\sin2x\right]_{\pi/6}^{\pi/2}\]

Answer 18

But whatev XD

Answer 19

... :> hehe? Sigh. I shall try to use less XD

Answer 20

This notation is very confusing by the way: \(\large\rm \sin2(x)\)
Try to stop doing that.

The 2 is part of the angle, so, if you want to use brackets,
write it like this: \(\large\rm \sin(2x)\)

Answer 21

ah gotcha.

Answer 22

Let's factor the negative out of the last piece,\[\large\rm \left[\sin2x+2\cos x\right]_0^{\pi/6}-\left[2\cos x+\sin2x\right]_{\pi/6}^{\pi/2}\]And then we just have to plug a bunch of stuff in, ya? :d

Answer 23

can
2cos(pi/6) become cos(pi/3) ?

Answer 24

sin(2x) will become sin(pi/3) when we plug in x=pi/6
but, no, not for the cosines :)
can't bring the 2 inside.

Answer 25

ahh ok
\[[\sin \frac{ \pi }{ 3 }+2\cos \frac{ \pi }{ 6 }-0-2\cos0]-[2\cos \frac{ \pi }{ 2 }+\sin \pi - 2\cos \frac{ \pi }{ 6 }-\sin \frac{ \pi }{ 3 }]\]

Answer 26

does that..look right?

Answer 27

Yes, perfect.

Answer 28

I would recommend...
convert to 0's and 1's anywhere that you're able to.
But wait on the other "special values", first combine like-terms if you're able.

Answer 29

\[[\frac{ \sqrt3 }{ 2 }+\sqrt3-0-1]-[0+0-\sqrt3-\frac{ \sqrt3 }{ 2 }]\]

Answer 30

didn't combine them yet, just to make sure i've got all those conversions correctly xD

Answer 31

Hmm your cosines gave you sqrt(3)'s? Hmmm

Answer 32

well 2*sqrt3/2

Answer 33

Oh I'm being silly :) nah you're good

Answer 34

Oh and also, in the first set of stuff,
the 2cos0 becomes 2*1, ya?

Answer 35

oo yes. did I put 1 for that? oopss

Answer 36

\[\frac{ 2\sqrt3 }{2 }+2\sqrt3-2\]\[\sqrt3+2\sqrt3-2\]\[=3\sqrt3-2\]

Answer 37

yay good job \c:/
and wolfram agrees, cool

Answer 38

haha yay!!! wolfram approved +zepdrix approved = green check mark xD

Answer 39

lol :D \(\huge\sf \color{green}{\checkmark}\)

Answer 40

YUSS

Answer 41

thank you quite much ^-^

Answer 42

np