Can someone look over my work?

- anonymous

Can someone look over my work?

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- anonymous

1. https://gyazo.com/0be194cb26fc09186e456b3040ae26bd
2. https://gyazo.com/8b9838a8f395e2f35cbab2deefbe0a64
3. https://gyazo.com/da6606ba2eb37743c14b793b253216e9
4. https://gyazo.com/9ada330db508c67ed1783e28ff35abfc
5. https://gyazo.com/69c59efbf2259744e9cdb2533723a917

- anonymous

@zepdrix @SolomonZelman @IrishBoy123 @Zarkon

- anonymous

Quite lost on what to do for 5.

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## More answers

- anonymous

Will probably post that one separately.

- IrishBoy123

start with \(a(t) = 40 \)
so \(v(t) = \int dt \; a(t) = 40t + \alpha\)
you also have \(v(0) = -20\) so \(\alpha = ????\)
just follow the clues that way.....

- anonymous

Ok, let me give it a go. Were you able to look at my other problems to see if I was on the right track?

- anonymous

is \[\alpha \] supposed to be C?

- IrishBoy123

it's a constant

- anonymous

@IrishBoy123 I can't seem to get it, I am stuck at
\[v(0)=-80+\alpha \]

- anonymous

What would I input for v(0) to solve? 10?

- anonymous

@PeterPan

- anonymous

That sure is a lot ^^
Are you like... in college already? haha

- anonymous

I wish, all this because me and my calc teacher got into an argument so he failed me. Now I gotta do it online :( then another time in college.

- terenzreignz

I'd have an easier time at this if we go at it one at a time...
Just for clarification, which problem are we currently working on? ^^

- anonymous

Let's start with the first.

- terenzreignz

Here goes nothing...
It seems the integral is a simple \[\Large \int_0^4\sin^{-1}\left(\frac{x}{4}\right)dx\]
And this much, you got, it seems...
Sorry if I'm starting over, I'm also not good at continuing someone else's solution, it puts me off-rhythm, you know how it is ^^

- anonymous

That's fine

- terenzreignz

For simplicity, you used substitution, letting \(\large m=\frac x4\) so that \(\large dm = \frac14dx\)

- anonymous

Yes, then I used integration by parts to integrate the function which I checked using wolfram

- terenzreignz

Can't go wrong with Wolfam :D
Were you correct according to it?

- anonymous

Yes

- anonymous

I then plugged in the endpoints and checked my answer using my graphing calculator

- anonymous

Which was right

- terenzreignz

Your calculator said you were right :D
Why defer to me, who's subject to human error? HAHA

- anonymous

I honestly just wanted to check my methods since I know the work should be fine if the method is correct.

- anonymous

LOOL

- anonymous

So we know #1 is right, and we worked out #2 (last post) so all we have left are 3-5

- terenzreignz

These people can be such sadists -_-

- terenzreignz

Well, actually, no.
#3 is pretty easy.
Any idea how to set up the bloody integral? :D

- terenzreignz

Here.
Well, say SOMETHING :D

##### 1 Attachment

- anonymous

So sorry, went to use the restroom.

- anonymous

I have already set up the integral in my work. If you didn't see it I can type it out for you.

- terenzreignz

Please do... if it's not too much trouble D:

- anonymous

Np

- anonymous

\[\int\limits_{0}^{1}(x-x^2)^2dx\]

- anonymous

Then I expanded

- anonymous

\[\int\limits_{0}^{1} x^4-2x^3+x^2dx\]

- anonymous

Took the integral and solved.

- terenzreignz

And knowing you, solving the actual integral was a no-brainer, yes? :D
Time to move on to #4?

- anonymous

Yessir hehe. I'm guessing I set it up correctly?

- terenzreignz

What? #3?
Yes.

- anonymous

Yay! :)

- anonymous

Alright #4.

- terenzreignz

As for #4, you'll have to tell me what the Mean Value Theorem for Integrals is in this case... there are two of them.

- anonymous

There is?

- anonymous

I wasn't given a case :(

- anonymous

Unless I'm supposed to figure that out

- anonymous

Cause I didn't, all I did was find slope and then set it equal to the derivative to solve for x

- terenzreignz

Well, that seems to be the normal Mean Value Theorem... as in, the one involving derivatives :D
But we need the Mean Value Theorem for Integrals here...

- terenzreignz

Anyway, yes, the mean value theorem for integrals does apply, now, why don't you start by integrating the function from -1 to 1? ^^

- terenzreignz

For your perusal:
http://www.mathwords.com/m/mean_value_theorem_integrals.htm

- anonymous

Should I erase what I had previously done?

- anonymous

Or does it apply in some way?

- terenzreignz

I didn't see what you had previously done D:

- terenzreignz

I'm sorry T.T
I don't have enough attention span to look through notes D:
That's why I never took any haha

- anonymous

I never took any either

- anonymous

But let me write it out

- anonymous

I found slope

- terenzreignz

oh, slope?
slope has nothing to do with this :D

- anonymous

\[\frac{ f(1)-f(-1)}{ 1+1}\]

- anonymous

Oh ok then

- anonymous

I'll start erasing :(

- terenzreignz

Integrate the function from -1 to 1 and what do you get?

- anonymous

I got 0

- anonymous

Site just lagged for me

- terenzreignz

Me too. It happens.
Okay, great ^^
The MVT for integrals says that for any continuous function, on an integral I, there is a point c in I such that f(c) = the integral.
So... what value/s from [-1 , 1] give 0 when you run them through f(x)?

- terenzreignz

Just solve for f(x) = 0 for the solutions that lie within [-1 , 1]

- anonymous

Site keeps glitching on me :( I'm doing that now.

- anonymous

I got x=0,4,-4 only one in the interval is 0

- anonymous

So final answer is x=0

- terenzreignz

Yup ^^

- anonymous

Great. Now last one.

- terenzreignz

Seems pretty straightforward to me :)

- anonymous

I tried following IrishBoy123's steps but I can't seem to get it.

- anonymous

I can't seem to solve for C

- terenzreignz

Then let's start over

- anonymous

\[\int\limits_{}^{} 40dt\]

- terenzreignz

With you so far... the answer should be
40t + C

- anonymous

\[= 40t\]

- anonymous

Yup

- anonymous

+C****

- terenzreignz

So... initial velocity is -20
It means when t = 0, the whole velocity should be -20.
40(0) + C = -20
Solve it... lol

- anonymous

Ohhhhhhhh I'm an idiot.

- terenzreignz

I'm Kurt :>

- anonymous

This is what I did \[v(0)=40(-20)+C\]

- anonymous

lol

- anonymous

Hi Kurt

- terenzreignz

That's how we learn :D

- terenzreignz

Okay, so I'm sure you've realised as I have that the velocity at time t would be
40t - 20
Now integrate this again.

- anonymous

Trial and error is the best way.

- terenzreignz

Got it? :D

- anonymous

\[20t^2-20t+C\]

- anonymous

Now solve for C again? Using initial position?

- terenzreignz

Yup :)
Gotten the hang of it, I see...

- anonymous

Hell yeah I did

- anonymous

Final answer is 10?

- terenzreignz

Finish it ^^

- anonymous

Plug in 10 to find s(t)?

- terenzreignz

Uhh... no.
Just set t=0 and have it equal to 10
And find C.
You complicate things too much :P

- anonymous

\[s(t)=20t^2-20t+10\]

- anonymous

Was C not 10?

- terenzreignz

C was 10.
It just wasn't accurate to say the *final answer* is 10.
The final answer is s(t), as you have so posted.
Well done :D

- anonymous

Hey this is math not english :p

- anonymous

Thanks for your help.

- anonymous

Probably gonna have more questions in like 10 minutes. I'm trying to finish the course by today

- terenzreignz

Hah... I'm so awesome :>

- anonymous

Indeed

- anonymous

Should I make one large post for the assignment or each question it's on post?

- anonymous

own*************

- terenzreignz

uhm one per post would be nice.
less lag somewhat

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