Question

Can someone look over my work?

Answer 1

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

Answer 2

@zepdrix @SolomonZelman @IrishBoy123 @Zarkon

Answer 3

Quite lost on what to do for 5.

Answer 4

Will probably post that one separately.

Answer 5

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.....

Answer 6

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?

Answer 7

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

Answer 8

it's a constant

Answer 9

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

Answer 10

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

Answer 11

@PeterPan

Answer 12

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

Answer 13

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.

Answer 14

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? ^^

Answer 15

Let's start with the first.

Answer 16

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 ^^

Answer 17

That's fine

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Yes

Answer 22

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

Answer 23

Which was right

Answer 24

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

Answer 25

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

Answer 26

LOOL

Answer 27

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

Answer 28

These people can be such sadists -_-

Answer 29

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

Answer 30

Here.
Well, say SOMETHING :D

Answer 31

So sorry, went to use the restroom.

Answer 32

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

Answer 33

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

Answer 34

Np

Answer 35

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

Answer 36

Then I expanded

Answer 37

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

Answer 38

Took the integral and solved.

Answer 39

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

Answer 40

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

Answer 41

What? #3?
Yes.

Answer 42

Yay! :)

Answer 43

Alright #4.

Answer 44

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.

Answer 45

There is?

Answer 46

I wasn't given a case :(

Answer 47

Unless I'm supposed to figure that out

Answer 48

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

Answer 49

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...

Answer 50

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

Answer 51

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

Answer 52

Should I erase what I had previously done?

Answer 53

Or does it apply in some way?

Answer 54

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

Answer 55

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

Answer 56

I never took any either

Answer 57

But let me write it out

Answer 58

I found slope

Answer 59

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

Answer 60

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

Answer 61

Oh ok then

Answer 62

I'll start erasing :(

Answer 63

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

Answer 64

I got 0

Answer 65

Site just lagged for me

Answer 66

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)?

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

So final answer is x=0

Answer 71

Yup ^^

Answer 72

Great. Now last one.

Answer 73

Seems pretty straightforward to me :)

Answer 74

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

Answer 75

I can't seem to solve for C

Answer 76

Then let's start over

Answer 77

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

Answer 78

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

Answer 79

\[= 40t\]

Answer 80

Yup

Answer 81

+C****

Answer 82

So... initial velocity is -20
It means when t = 0, the whole velocity should be -20.

40(0) + C = -20

Solve it... lol

Answer 83

Ohhhhhhhh I'm an idiot.

Answer 84

I'm Kurt :>

Answer 85

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

Answer 86

lol

Answer 87

Hi Kurt

Answer 88

That's how we learn :D

Answer 89

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.

Answer 90

Trial and error is the best way.

Answer 91

Got it? :D

Answer 92

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

Answer 93

Now solve for C again? Using initial position?

Answer 94

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

Answer 95

Hell yeah I did

Answer 96

Final answer is 10?

Answer 97

Finish it ^^