Question

Find the area between two curves y=e^2x and y=sin2x where 0 < x< (pi/2)
- -

Answer 1

is that e^(2x) or e^2 * x
and

sin^2(x) or sin(2x)

Answer 2

e^(2x) and sin(2x)

Answer 3

The one problem is seem to be having is finding out when e^(2x)-sin(2x)=0
The rest I can do.

Answer 4

Hello?

Answer 5

you'll need to use a graphing calculator to find the approximate solutions to e^(2x)-sin(2x)=0

Answer 6

I don't think it's possible to solve e^(2x)-sin(2x)=0 without using a calculator

Answer 7

something wrong here

Answer 8

they intersect a bunch of places to the left of zero

Answer 9

http://www.wolframalpha.com/input/?i=e^%282x%29%2Csin%282x%29+x+from+-10+to+0

Answer 10

The calculator is giving me the wrong area, its going over regions that are not suppose to be shaded.

Answer 11

also the answer has to be within 0 and pi/2

Answer 12

sine is periodic, and \(e^{2x}\) has a horizontal asymptote at the x axis, so they intersect infinitely often

Answer 13

which is why I was confused on what to do since there is no real answer for e^2x-sin2x=0. Because if I get the answer to that then I know how to set up the second integral

Answer 14

you are looking for this green area (see attached image)

the red curve is the function y = sin(2x)
the blue curve is y = e^(2x)

the bounds on x are 0 < x < pi/2

Answer 15

so you just need to compute the integrals from 0 to pi/2 (for the two functions) and subtract

Answer 16

you could try this
http://www.wolframalpha.com/input/?i=integral+0+to+2pi+%28e^%282x%29-sin%282x%29%29

Answer 17

Yes, when I do that on my TI-84 the area beneath the curve in between x=0 and x=1 is being shaded as well

Answer 18

I keep getting the answer to 10. something when its suppose to be 8. something

Answer 19

it's not between x = 0 and x = 1

it's between x = 0 and x = pi/2

Answer 20

okay thanks

Answer 21

half e to the pi minus 3 halves

Answer 22

\[\frac{ e^\pi }{ 2 }-\frac{ 3 }{ 2 }\]

Answer 23

how do I show my work for this?

Answer 24

Calculate 2 definite integrals for x from zero to half pi

Answer 25

Yeah I understand that one goes from pi/2 to 0. And the other goes from pi/2 to what number?

Answer 26

\[\int\limits_{0}^{\pi/2}e^{2x}dx - \int\limits_{0}^{\pi/2}\sin(2x)dx\]

Answer 27

that 's what I'm having trouble with, in other to find that other number I have to find the answer to this equation: e^2x-sin2x=0. But since there is no answer to this , that why I'm confused

Answer 28

are you sure thats how you do it? because this is why my professor told us to do

Answer 29

the two functions don't intersect on that interval

Answer 30

you only do e^2x-sin2x=0 when there are points of intersection (between x = 0 and x = pi/2)

Answer 31

so you know which function is above the other (and where this occurs)

Answer 32

\[\int\limits_{0}^{\pi/2}e^2x-(\sin(2x)\]

Answer 33

i thought thats how I'm suppose to do this problem

Answer 34

and the second equation you would add to the previous one

Answer 35

you have it correct, you can break it up like dan is showing

Answer 36

\[\int\limits_{\pi/2}^{don't know this}\]

Answer 37

it's from 0 to pi/2, so you do know the upper piece

Answer 38

the graph shows no intersections between x = 0 and x = pi/2

Answer 39

but the answer that I am getting 10.1 is not right the answer should be 8. something

Answer 40

\[\frac{ e^{2x} }{ 2 }-\frac{- \cos(2x) }{ 2 }\]

that is what the integrals evaluate to, now just use upper bound pi/2 and lower bound zero, to calculate the value for the definate integral

Answer 41

I'm getting 10.07035 roughly

not sure how it's 8 something

Answer 42

yeah that's what I got too but my professor said the answer is 8 something

Answer 43

something seems missing if that's the case

Answer 44

not sure what

Answer 45

yeah that's why I thought that there should be a seccond equation

Answer 46

can you post a screenshot of the full question?

Answer 47

here ill show you an example from my online homework

Answer 48

yeah I will

Answer 49

You get 8.167 if you integrate from x=1 to x=pi/2 and add them instead of subtract. lol i dont know what you looking for.

Answer 50

im so sorry, for the confusion. But here is the question

Answer 51

its problem 8b

Answer 52

yeah the answer is 10.07035 (approximately)

there is either a typo on the sheet or the professor is mistaken

Answer 53

oh okay! Thank for so much for your help! I appreciate it greatly. And sorry for not understanding that quickly.

Answer 54

yeah.

The first one 8a uses that process you described, setting the two functions equal to find the bounds for integration, -2 to 3

Answer 55

that's ok, you picked it up just fine

you're welcome