anonymous
  • anonymous
Can someone look over my work?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
@zepdrix @SolomonZelman @IrishBoy123 @Zarkon
anonymous
  • anonymous
Quite lost on what to do for 5.

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anonymous
  • anonymous
Will probably post that one separately.
IrishBoy123
  • 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
  • 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
  • anonymous
is \[\alpha \] supposed to be C?
IrishBoy123
  • IrishBoy123
it's a constant
anonymous
  • anonymous
@IrishBoy123 I can't seem to get it, I am stuck at \[v(0)=-80+\alpha \]
anonymous
  • anonymous
What would I input for v(0) to solve? 10?
anonymous
  • anonymous
@PeterPan
anonymous
  • anonymous
That sure is a lot ^^ Are you like... in college already? haha
anonymous
  • 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
  • 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
  • anonymous
Let's start with the first.
terenzreignz
  • 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
  • anonymous
That's fine
terenzreignz
  • terenzreignz
For simplicity, you used substitution, letting \(\large m=\frac x4\) so that \(\large dm = \frac14dx\)
anonymous
  • anonymous
Yes, then I used integration by parts to integrate the function which I checked using wolfram
terenzreignz
  • terenzreignz
Can't go wrong with Wolfam :D Were you correct according to it?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
I then plugged in the endpoints and checked my answer using my graphing calculator
anonymous
  • anonymous
Which was right
terenzreignz
  • terenzreignz
Your calculator said you were right :D Why defer to me, who's subject to human error? HAHA
anonymous
  • anonymous
I honestly just wanted to check my methods since I know the work should be fine if the method is correct.
anonymous
  • anonymous
LOOL
anonymous
  • anonymous
So we know #1 is right, and we worked out #2 (last post) so all we have left are 3-5
terenzreignz
  • terenzreignz
These people can be such sadists -_-
terenzreignz
  • terenzreignz
Well, actually, no. #3 is pretty easy. Any idea how to set up the bloody integral? :D
terenzreignz
  • terenzreignz
Here. Well, say SOMETHING :D
1 Attachment
anonymous
  • anonymous
So sorry, went to use the restroom.
anonymous
  • 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
  • terenzreignz
Please do... if it's not too much trouble D:
anonymous
  • anonymous
Np
anonymous
  • anonymous
\[\int\limits_{0}^{1}(x-x^2)^2dx\]
anonymous
  • anonymous
Then I expanded
anonymous
  • anonymous
\[\int\limits_{0}^{1} x^4-2x^3+x^2dx\]
anonymous
  • anonymous
Took the integral and solved.
terenzreignz
  • terenzreignz
And knowing you, solving the actual integral was a no-brainer, yes? :D Time to move on to #4?
anonymous
  • anonymous
Yessir hehe. I'm guessing I set it up correctly?
terenzreignz
  • terenzreignz
What? #3? Yes.
anonymous
  • anonymous
Yay! :)
anonymous
  • anonymous
Alright #4.
terenzreignz
  • 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
  • anonymous
There is?
anonymous
  • anonymous
I wasn't given a case :(
anonymous
  • anonymous
Unless I'm supposed to figure that out
anonymous
  • anonymous
Cause I didn't, all I did was find slope and then set it equal to the derivative to solve for x
terenzreignz
  • 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
  • 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
  • terenzreignz
For your perusal: http://www.mathwords.com/m/mean_value_theorem_integrals.htm
anonymous
  • anonymous
Should I erase what I had previously done?
anonymous
  • anonymous
Or does it apply in some way?
terenzreignz
  • terenzreignz
I didn't see what you had previously done D:
terenzreignz
  • 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
  • anonymous
I never took any either
anonymous
  • anonymous
But let me write it out
anonymous
  • anonymous
I found slope
terenzreignz
  • terenzreignz
oh, slope? slope has nothing to do with this :D
anonymous
  • anonymous
\[\frac{ f(1)-f(-1)}{ 1+1}\]
anonymous
  • anonymous
Oh ok then
anonymous
  • anonymous
I'll start erasing :(
terenzreignz
  • terenzreignz
Integrate the function from -1 to 1 and what do you get?
anonymous
  • anonymous
I got 0
anonymous
  • anonymous
Site just lagged for me
terenzreignz
  • 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
  • terenzreignz
Just solve for f(x) = 0 for the solutions that lie within [-1 , 1]
anonymous
  • anonymous
Site keeps glitching on me :( I'm doing that now.
anonymous
  • anonymous
I got x=0,4,-4 only one in the interval is 0
anonymous
  • anonymous
So final answer is x=0
terenzreignz
  • terenzreignz
Yup ^^
anonymous
  • anonymous
Great. Now last one.
terenzreignz
  • terenzreignz
Seems pretty straightforward to me :)
anonymous
  • anonymous
I tried following IrishBoy123's steps but I can't seem to get it.
anonymous
  • anonymous
I can't seem to solve for C
terenzreignz
  • terenzreignz
Then let's start over
anonymous
  • anonymous
\[\int\limits_{}^{} 40dt\]
terenzreignz
  • terenzreignz
With you so far... the answer should be 40t + C
anonymous
  • anonymous
\[= 40t\]
anonymous
  • anonymous
Yup
anonymous
  • anonymous
+C****
terenzreignz
  • 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
  • anonymous
Ohhhhhhhh I'm an idiot.
terenzreignz
  • terenzreignz
I'm Kurt :>
anonymous
  • anonymous
This is what I did \[v(0)=40(-20)+C\]
anonymous
  • anonymous
lol
anonymous
  • anonymous
Hi Kurt
terenzreignz
  • terenzreignz
That's how we learn :D
terenzreignz
  • 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
  • anonymous
Trial and error is the best way.
terenzreignz
  • terenzreignz
Got it? :D
anonymous
  • anonymous
\[20t^2-20t+C\]
anonymous
  • anonymous
Now solve for C again? Using initial position?
terenzreignz
  • terenzreignz
Yup :) Gotten the hang of it, I see...
anonymous
  • anonymous
Hell yeah I did
anonymous
  • anonymous
Final answer is 10?
terenzreignz
  • terenzreignz
Finish it ^^
anonymous
  • anonymous
Plug in 10 to find s(t)?
terenzreignz
  • terenzreignz
Uhh... no. Just set t=0 and have it equal to 10 And find C. You complicate things too much :P
anonymous
  • anonymous
\[s(t)=20t^2-20t+10\]
anonymous
  • anonymous
Was C not 10?
terenzreignz
  • 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
  • anonymous
Hey this is math not english :p
anonymous
  • anonymous
Thanks for your help.
anonymous
  • anonymous
Probably gonna have more questions in like 10 minutes. I'm trying to finish the course by today
terenzreignz
  • terenzreignz
Hah... I'm so awesome :>
anonymous
  • anonymous
Indeed
anonymous
  • anonymous
Should I make one large post for the assignment or each question it's on post?
anonymous
  • anonymous
own*************
terenzreignz
  • terenzreignz
uhm one per post would be nice. less lag somewhat

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