Question

Find the exact area of the region that lies under the curve given by
y=2sin(x)−sin(2x),0≤x≤π

Answer 1

I know to change the problem into an integral notation
\[\int\limits_{0}^{\pi} 2\sin(x)-\sin(2x)\]
u=2x
du=2 dx
dx=1/2 du

the question is what do i do with the other x?

Answer 2

If you want to, you can split up the integral, and only apply the `u sub` to the second one.\[\int\limits\limits_{0}^{\pi} 2\sin(x)-\sin(2x) \quad = \quad \int\limits\limits_{0}^{\pi} 2\sin(x)dx-\int\limits\limits_{0}^{\pi} \sin(2x)dx\]

Answer 3

why not the first one too?

Answer 4

It doesn't look like you need to apply it to the first one :O unless it's confusing you.

Answer 5

and make the u=x
and du=1 dx
and dx= du

Answer 6

yes it is confusing me a lot

Answer 7

Sure you can do that if you want :)
Two separate u-subs so both answers are in terms of u.
That will work just fine ^^

Answer 8

ok ill try that thanks for your help

Answer 9

When you get more comfortable with integrals, you'll be able to skip the `u sub` on problems like this.

Because it's really just thinking of the chain rule in a different way.
If we wanted to take the `derivative` of \(\sin(2x)\) a \(2\) would pop out due to the chain rule right?
Well, since we're integrating it, the opposite will happen, a \(\dfrac{1}{2}\) will pop out, so to speak :D

Answer 10

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

Answer 11

yep looks good.

Answer 12

now should i combine them so they are
2sin(u)-sin(u)1/2 du?????

Answer 13

and the 2's cancel

Answer 14

Well we can't combine them since it's a `definite integral` and we applied two DIFFERENT substitutions.
They now have different limits of integrations.\[\huge \int\limits\limits_{x=0}^{\pi} 2 \sin(u)du-\int\limits\limits_{x=0}^{\pi}\sin(u)\frac{ 1 }{2 }du\]

Answer 15

See how our limits are in terms of `x`?
We need limits in terms of `u`.

Answer 16

i got there

Answer 17

I just don't know what to do next

Answer 18

I think the limits would change on the second one yes?\[\huge \int\limits\limits\limits_{x=0}^{\pi} 2 \sin(u)du-\int\limits\limits\limits_{\color{royalblue}{u=0}}^{\color{royalblue}{2\pi}}\sin(u)\frac{ 1 }{2 }du\]

Answer 19

This is a really bad way to do this problem :( Hmmm.

Answer 20

We really should start over and do it differently.
But if you want to continue doing it this way, then that's ok, we can write the integrals like this, pulling the constants outside.\[\huge 2\int\limits\limits_{x=0}^{\pi} \sin(u)du-\frac{1}{2}\int\limits\limits_{\color{royalblue}{u=0}}^{\color{royalblue}{2\pi}}\sin(u)du\]Then we can integrate them separately from here.

Answer 21

Woops, that first one should be u=0 at the bottom. my bad.

Answer 22

Uh oh did I confuse you? Is your brain gonna esplode? :c

Answer 23

I hate when that happens... brains everywhere...

Answer 24

no not yet im trying to finish the problem

Answer 25

haha nope not me

Answer 26

is the right answer -3?

Answer 27

I think the answer is +4.
Lemme check my work again.

Answer 28

Yah it's 4. Did we run into a little mistake somewhere? <:o

Answer 29

yes i have an idea to where that happened thanks for the help

Answer 30

oh ok cool c: