Question

Finding the ARea of the Region

f(x)= 2sinx + sin2x
y=0

Answer 1

\[0 \le x \le \pi\] interval

Answer 2

I have this so far

\[\int\limits_{0}^{\pi} (2\sin x + \sin2x) dx\]

Answer 3

I have to integrate...so I got this...

\[-2\cos x + ??? \]

Answer 4

seems to me you did a good start

Answer 5

I don't get how to integrate sin2x ? i Know sinx = -cosx but what if it has 2 in there?

Answer 6

\[\int_{0}^{\pi}2\sin x dx+\int_{0}^{\pi}\sin 2x dx=-2\cos|_{0}^{\pi}+\int2\sin x\cos xdx\]

Answer 7

can you intgerate that sin 2x now

Answer 8

the first one is easy we are done
now focus on int (sin 2x ) which is the same as 2int(cosx sinx )dx

Answer 9

oh okay...

Answer 10

so does the 2 come to the front of sin ?

will it be -2cosx ?

Answer 11

xapproachesinfinity used the identity

sin(2x) = 2*sin(x)*cos(x)

Answer 12

yes! exactly like jim said

Answer 13

although now that I look at it, it's not a good move to use it

Answer 14

whoops :/

Answer 15

use u-substitution

let u = 2x

Answer 16

you can also use substitution either way work

Answer 17

oh true, you can

Answer 18

i like the u substitution

Answer 19

if you let u = 2x, then du = 2*dx ---> dx = du/2

Answer 20

i have 1/2 = du

Answer 21

i am not too great at trig sorry @xapproachesinfinity

Answer 22

replace dx be du/2
so to get int(sinudu/2)

Answer 23

yep!! but then the 1/2 goes infront of the integral right ?

Answer 24

\[\int_{0}^{\pi}\sin u\frac{du}{2}\]
1/2 is constant so you can factor it out
\[\frac{1}{2}\int_{0}^{\pi}\sin u du\]

Answer 25

yes correct 1/2 goes in front :)

Answer 26

oh the limits change though

Answer 27

xapproachesinfinity, the limits aren't going to be 0 and pi

if u = 2x, then

when x = 0 ---> u = 2x = 2(0) = 0
when x = pi ---> u = 2(pi) = 2pi

so the limits are now from 0 to 2pi

Answer 28

not 0 to pi

Answer 29

yeah you caught it as I was writing

Answer 30

yea i just caught that :) thanks

Answer 31

or you can just revert back to x (by plugging u = 2x back in) and not worry about changing the limits

Answer 32

after you integrate

Answer 33

yea that's another way to do it
well as long as the asker is with us lol

Answer 34

did you understand the u sub we just did?

Answer 35

yes but i'm stuck on putting the u and du back in now before I integrate

Answer 36

all you need to do now is integrate \[1/2\int\sin udu\]

Answer 37

forgot the limits for now
we gonna go back to it

Answer 38

\[\frac{ 1 }{ 2 }\int\limits_{0}^{\pi} du * u ??\]

Answer 39

no where did you get u
here is what we did
let u=2x du=2dx
so dx=1/2 du
\[\int\sin2xdx=\int\sin( u)\frac{du}{2}=1/2\int\sin( u)du\]

Answer 40

just repeated the us sub so to make sure you got it

Answer 41

now it is an easy integral
just think what function gives sin x when differentiate it?

Answer 42

ok here is what i have

\[\int\limits_{0}^{\pi} \sin(u) du\]

Answer 43

o h crap i forgot the 1/2

Answer 44

no not 0 to pi
forgot those limits for now just integrate

Answer 45

just find the anti derivative first

Answer 46

\[\frac{ 1 }{ 2 } \int\limits_{0}^{\pi} \sin(u) du\]

Answer 47

but not 0 to pi okay
that's why i said forgot the limits for now

Answer 48

that's an error in my side as jim mentioned

Answer 49

okay so are we now just finding the antiderv of sin2x right ? or the whole thing ?

Answer 50

we find the antiderivative of sin u
which will give us the antiderivative of sin 2x

Answer 51

not the whole think of course one part is done already

Answer 52

thing*

Answer 53

oh okay!! i got \[\frac{ 1 }{ 2 } -\cos2x\]

Answer 54

can it be -1/2cos2x ?

Answer 55

okay but you forgot 1/2 that was attached to it

Answer 56

that is good but you need to multiply by 1/2

Answer 57

-1/2cosx is that right ?

Answer 58

\[1/2\int \sin udu=-1/2\cos u=-1/2\cos (2x)\]
sorry you got the correct one

Answer 59

no the whole thing
\[=-2\cos x|_{0}^{\pi}+(-1/2 \cos 2x|_{0}^{\pi}\]

Answer 60

just evaluate now

Answer 61

sorry my attention is little spread heheh

Answer 62

okay im going to evaluate

Answer 63

ok

Answer 64

-2cosx - 1/2cosx

Answer 65

2-1/2 = 3/2

Answer 66

i plugged in pi

Answer 67

hmm you didn't do it right
\[(-2\cos\pi+2\cos 0)+(-1/2 \cos 2\pi+1/2\cos0)\]
is this what you did

Answer 68

you should have 2+2-1/2 +1/2=4

Answer 69

oh no i didn't do that...but now i will !

Answer 70

okay

Answer 71

well how do you evaluate the integral \[\int _{a}^{b}f(x)dx=(F(x)|_{a}^{b}=F(b)-F(a)\]

Answer 72

i got zero ?? what's going on :(

Answer 73

well check again

Answer 74

okay i see, I subtracted instead of adding

Answer 75

what did you get

Answer 76

still 0

Answer 77

hahaha
show me your work

Answer 78

this is what i have
-2cos(0) + 1/2cos(0) + -2cos(pi) +1/2cos(pi)

Answer 79

hmm are you sure check what i wrote?

Answer 80

where is 2pi in there

Answer 81

now i see the error!!
okay....here goes

Answer 82

why did we change the interval? i know it happened earlier, but why ?

Answer 83

http://www.wolframalpha.com/input/?i=int_%7B0%7D%5E%7Bpi%7D%282sinx%2Bsin2x%29dx
check here the same result i got

Answer 84

-2cos(pi) +2cos(0) -1/2cos(2pi) + 1/2cos(pi)

Answer 85

i didn't change the limits actually still 0 to pi

Answer 86

i saw the wolfframalpha, ok but why did we plug in 2pi ?

Answer 87

well that's why i told you to not think about them
jim just mentioned and error i made when i put the limits of the integral
when i did U sub
but since we went back to x again then we should not change the limits

Answer 88

not 2pi
we plug is pi
the function is cos 2x

Answer 89

but i plugged in 2pi on -1/2cos(2pi)....

Answer 90

the integral gave us -1/2 cos(2x) remember?

Answer 91

so if we plug pi we get -1/2 cos (2 pi)

Answer 92

ohhhh okay!!! i get it now!

Answer 93

Okay good!

Answer 94

I got 3 now, I am getting close lol !

Answer 95

okay perhaps you forgot something there just check

Answer 96

the -1/2 + 1/2 cancel out ....so I should get 4, an error in my signs

Answer 97

yup exactly :)

Answer 98

good job