Question

Anyone up for a challenge? Find the area bounded by these three curves: y=4-4x; y=2xcos(x^2); y=sin(x)cos^2(x) in the first quadrant. Use the calculator to get the intersection points...I just can't seem to get the same answer twice! Algebra issues...

Answer 1

It's the limits of integration, you know plugging everything in that's getting me. Let me know the answers you get.

Answer 2

from the picture it looks like the left hand endpoint is where
\[2x\cos^2(x)=\sin(x)\cos^2(x)\] i.e.
\[x=0\]

Answer 3

Yes, I agree.

Answer 4

The endpoints I get for the other two intersections are .6928 and 1.237

Answer 5

( I just plugged the functions into my graphing calculator and evaluated the intersections where it seemed that the graphs overlapped.)

Answer 6

then next one is where
\[4-4x=2x\cos^2(x)\] wolfram gives me
\[x= 0.708523315642791...\]

Answer 7

Really!
I should have used wolfram. Let me try it again with new numbers.

Answer 8

http://www.wolframalpha.com/input/?i=y%3D4-4x+and++y%3D2xcos%28x^2%29and+y%3Dsin%28x%29cos^2%28x%29
http://www.wolframalpha.com/input/?i=2xcos^%28x%29+%3D+4-4x

Answer 9

still have to find where
\[4-4x=\sin(x)\cos^2(x)\]

Answer 10

We can do that the same way--wolfram, right?

Answer 11

yeah i got
\[0.928113\]
http://www.wolframalpha.com/input/?i=4-4x%3Dsin%28x%29cos^2%28x%29

Answer 12

Hmm. The first one is actually 2xcos(x^2)

Answer 13

It's not 2xcos^2(x). So, 4-4x=2xcos(x^2) = .6928

Answer 14

So, 0, .6928, .9281

Answer 15

You found, accidentally 4-4x=2xcos^x(x) and got .7085, I think.

Answer 16

oops
i tried it again and got this
http://www.wolframalpha.com/input/?i=2xcos^2%28x%29+%3D+4-4x

Answer 17

No no! It's 2xcos(x^2). Not cosine squared of x, but cosine of x squared. ><

Answer 18

oh lord

Answer 19

.6828 it is

Answer 20

It's okay :P i just found out how to evaluate the definite integral on my calculator, so I might be able to get this now! Doing it by hand was TOO much.

Answer 21

But if you'd do it too so we can check answers...

Answer 22

forget doing it by hand

Answer 23

That'd be awesome.

Answer 24

OH I KNOW. It was just awful.

Answer 25

now we just have to figure out what to integrate

Answer 26

mhm. I split it up. So from 0 to .6928, you do 2xcos(x^2)-(4-4x)?

Answer 27

no, not from the picture
looks like you should do this
http://www.wolframalpha.com/input/?i=integral+0+to+0.692751+%282xcos^2%28x%29-sin%28x%29cos^2%28x%29%29

Answer 28

No. no. 2xcos(x^2)-sinxcos^2(x)

Answer 29

Yeah, what you said :P

Answer 30

whew

Answer 31

over that interval
\[2x\cos(x^2)>\sin(x)\cos^2(x)\]

Answer 32

Yes.

Answer 33

HEY SATELLITE! We got it right! It was an option! Nice! Thank you so much :)

Answer 34

wait!!

Answer 35

we still have to do the second part

Answer 36

Oh, I added the other side already

Answer 37

http://www.wolframalpha.com/input/?i=integral+0.692751+to+0.928113+%284-4x-sin%28x%29cos^2%28x%29%29+

Answer 38

.378 something is the anwer.

Answer 39

Don't worry :) I got it. thanks for the suggestion of using wolfram, though. It's much easier than typing all of this in the calculator.

Answer 40

yw but that is not what i get

Answer 41

i got .195.. for the first part, and .098 for the second part, but i could be wrong

Answer 42

If you add both together? The first section = approximately .208 and the second part = approximately .0983. You add them together to equal .378. Which was the correct answer.

Answer 43

but if you got the right answer great!

Answer 44

We differ on the first part.

Answer 45

Might have to do with rounding. I dunno. But thanks so much for the help :)

Answer 46

yw