Question

Sketch the region enclosed by the given curves.
y = 5 cos 4x, y = 5 − 5 cos 4x, 0 ≤ x ≤ π/4
Find it's area.

Answer 1

This is a graph of the region

Answer 2

I figured you had to go ahead and add the following integrations \[(\int\limits_{0}^{\Pi/12} (5\cos(4x)) - (5-5\cos(4x))dx ) + (\int\limits_{\Pi/12}^{\Pi/4} (5-5\cos(4x))-(5\cos(4x))dx)\]
but every time I try it, the answer I get is wrong. Help !

Answer 3

Hello @beeca_boo73 Welcome to Openstudy :)

Answer 4

Hi @Luigi0210 ! Thanks, I'm excited to discover this nifty site. Hopefully it'll be helpful. (:

Answer 5

@Psymon will assist you with your question.

Answer 6

Im already in 2 questions @Luigi0210

Answer 7

Btw if you have any questions or concerns about the site feel free to PM me, or anybody else with an A next to their name, hope you enjoy the site :)

Answer 8

Okay, I'll try and help.

Answer 9

Okay, thanks. (:
I'm just starting calc II and haven't had calc AB for a year and a half so I really can't remember how to do the integrations of trig functions in general.

Answer 10

Well okay, so you figured out where they intersect right?

Answer 11

Yeah, I did. They intersect at \[\Pi/12\]
I think I have the integration formula right I just don't know how to take the antiderivative of the cos functions.

Answer 12

\[\int\limits cosx dx=-sinx +C\]
If I remember correctly

Answer 13

@Loser66 We have a student here waiting to learn :3

Answer 14

calc 2

Answer 15

Yeah, this is calc 2 stuff. Still in one problem, haha

Answer 16

Yeah, calc II, semi rehash from calc AB.

Answer 17

@Psymon I thought you were busy?

Answer 18

So yeah, you need to determine a few things. the correct limits, the point of intersection and which function is higher and which function is lower. This will require two integrals because of us crossing and changing which function is higher. So before they intersect, which function is the higher of the two functions?

Answer 19

Nah, end the first integral at the intersection.

Answer 20

I think I see where you're going with this

Answer 21

Before they intersect, the 5cos(4x) is on top, correct ? So the integration would be 5cos(4x) - (5-5cos(4x))

Answer 22

Correct.

Answer 23

But then you have to swap them after that because 5cos(4x) is no longer on top.

Answer 24

Right. In which case you would do 5-5cos4x - 5cos4x for the second integral.

Answer 25

And you integrate from the intersection to pi/4

Answer 26

The first integral you integrate up until the intersection. Then you add the second integral with limits starting at the intersection and ending at pi/4

Answer 27

\[\int\limits_{0}^{intersection}5\cos4x - (5-5\cos4x)dx + \int\limits_{intersection}^{\frac{ \pi }{ 4 }}5-5\cos4x-(5\cos4x)dx\]

Answer 28

Right, I got that. I originally got pi/12 as the place they intersect. Was that correct or no?

Answer 29

Ill check.

Answer 30

Thanks for trying @Loser66 ! (:

Answer 31

Yes, pi/12 is correct. : )

Answer 32

Therefore we have this :P
\[\int\limits_{0}^{\frac{ \pi }{ 12 }}5\cos4x- (5-5\cos4x)dx + \int\limits_{\frac{ \pi }{ 12 }}^{\frac{ \pi }{ 4 }}5-5\cos4x-(5\cos4x)dx\]

Answer 33

Yeah. Okay, great, so my set up is right. Yay! Now my main problem is I suck at integrating trig functions. So, I began by simplifying the integration to
\[\int\limits_{0}^{\Pi/12}(10\cos(4x) - 5)dx + \int\limits_{\Pi/12}^{\Pi/4} (5-10\cos(4x))dx\]
Does that look right?

Answer 34

Yes it does.

Answer 35

Now let's start with the first integral. Because cos is only to the first power and we have no wandering x's, we can simply say this:
let u = 4x. therefore du = 4dx and dx = du/4.
\[\int\limits_{}^{}10\cos(u)*\frac{ du }{ 4 }\]

Answer 36

Okay. And can you pull the 10 to the front of the integral for that ?
So when you integrate for cosu you'd get.. sinu/4 ?

Answer 37

Well you can factor out 10/4, but don't forget to put it back. So yes, cosine integrates to sin u where u = 4x. So what we end up with is:
\[\frac{ 10\sin(u) }{ 4 }=\frac{ 5\sin(4x) }{ 2 } \]

Answer 38

Okay, so you'd have 5sin(4x)/4 minus \[\int\limits_{}^{}5dx\] which would be 5x so for the first integration you'd end up with 5sin(4x)/2 - 5x from 0 - pi/12

Answer 39

We're working on that @Loser66 xD and yes, exactly. So we cansolve that and then get to our 2nd integral :3

Answer 40

?

Answer 41

We have two "intervals" though.

Answer 42

sin(pi/3) is the square root of 2 over 2, right ?

Answer 43

sqrt(3)/2

Answer 44

Aw, crap, okay, lol

Answer 45

Lol, we'll see if we run into any issues @Loser66 Thank you for watching over our work, though xD

Answer 46

So the first integration would end up being \[5\sqrt{3}/4 - 5\sqrt{3}/12\]
?

Answer 47

no wait minus 5(pi)/12

Answer 48

Lol, you caught yourself. Correct. Now we can work on adding the result of the 2nd interval : )

Answer 49

Okay. 5 - 10cos(4x) so you'd separate it to the integral of 5dx, which would be 5x, minus the integral of 10cos(4x), which would end up being the same 5sin(4x)/2 right ?

Answer 50

Right, so we have 5x - 5sin(4x)/2 from pi/12 - pi/4

Answer 51

Okay and you input the pi/4 for all the x's first, get that value and subtract the value from the expression when pi/12 is input for the x's after, right ?

Answer 52

Yep. Same thing we did from the first integral f(pi/4) - f(pi/12)

Answer 53

so it would be (-5(pi)/4) - (5(sqrt 3)/4 - 5(pi)/12)

Answer 54

?

Answer 55

5p/4 is positive.

Answer 56

but I thought you'd subtract it from the 5sin(pi)/2 which would be zero ?

Answer 57

Oh I put it in backwards

Answer 58

Lol, no worries.

Answer 59

We should have this:
\[\frac{ 5\pi }{ 4 }-(\frac{ 5\pi }{ 12 }-\frac{ 5\sqrt{3} }{ 4 })\]

Answer 60

I was using the formula from the first integration, my bad. okay so positive 5(pi)/4)
so we'd get 5(pi)/6 - 5(sqrt 3)/4

Answer 61

And add that to the first integrations answer of 5(sqrt 3)/4 - 5(pi)/12

Answer 62

Yes.

Answer 63

\[\frac{ 5\sqrt{3} }{ 4 }-\frac{ 5\pi }{ 12 }+\frac{ 5\pi }{ 6 }+\frac{ 5\sqrt{3} }{ 4 } \]

Answer 64

so we'd get 5(sqrt3)/2 + 5(pi)/12 as the final answer ?

Answer 65

Yep xD

Answer 66

THANK THE LORD
You are a beautiful person, thanks so much. (:

Answer 67

Yep np ^_^