Question

Hiiii can someone help me with any of these three questions? ThAnKs!!

Answer 1

what are they?

Answer 2

Sorry i'm trying to attach it!

Answer 3

@Lolipop_Drum312

Answer 4

oh hold on

Answer 5

okay!

Answer 6

which of them do you know how to do?

Answer 7

its okay thanks anyways!

Answer 8

Hey Natasha! :)
To go backwards and find f, given f'', we'll need to anti-differentiate.
(Or integrate, I dunno if you've learned that term at this point.)
\[\Large\rm f''(x)=-4\sin(2x)\]

Answer 9

Oh there an integral in the next question XD so obviously you have haha

Answer 10

\[\Large\rm f'(x)=\int\limits -4\sin(2x)dx\]

Answer 11

haha yeah we have but it was a while ago! can you refresh my memory?

Answer 12

Umm, to deal with this.. you `can` apply a u-substitution if you're more comfortable with that.

\(\Large\rm u=2x\)
\(\Large\rm du=2dx\qquad\to\qquad \frac{1}{2}du=dx\)

Then your problem becomes:\[\Large\rm \int\limits -4\sin(u)\left(\frac{1}{2}du\right)\]

Answer 13

\[\Large\rm =-2\int\limits \sin(u)du\]Do you remember how to integrate sine? :o
derivative of sine is cosine, what do we get if we go backwards though? :)

Answer 14

-cos!

Answer 15

good good good, and now i see that i shouldn't had pulled the negative out front lol\[\Large\rm =-2(-\cos u)+c\]\[\Large\rm =2\cos(u)+c\]
Undoing our substitution,
\[\Large\rm f'(x)=2\cos(2x)+c\]

Answer 16

They gave us some `initial data`.
This allows us to deal with the constant c that shows up when integrating.
\(\Large\rm f'(0)=2\)

So if you plug in 0 for your x, and 2 for your "y", what do you get for c? :)

Answer 17

would c just be 2?

Answer 18

\[\Large\rm f'(x)=2\cos(2x)+c\]Umm let's check.
We'll plug in x=0, y=2,\[\Large\rm 2=2\cos(2\cdot0)+c\]\[\Large\rm 2=2\cos(0)+c\]Cosine of 0 gives us.. 1, ya? :)\[\Large\rm 2=2+c\]

Answer 19

okay yeah i wasnt sure if was that or 2! so just 0?

Answer 20

Ok great, so our f'(x) simplifies down a little bit since our c is zero.\[\Large\rm f'(x)=2\cos(2x)\]We need to integrate again to get from f' to f.

Answer 21

\[\Large\rm f(x)=\int\limits 2\cos(2x)dx\]

Answer 22

We can apply a u-sub AGAIN if you want -_-
But I would recommend trying to get used to this little shortcut. umm

Answer 23

Remember when you take a derivative, something like this,\[\Large\rm (\sin2x)'=2cos2x\]You get an extra 2 popping out because of the chain rule, yah?

Answer 24

\[\Large\rm \int\limits \cos2x~dx=\frac{1}{2}\sin2x\]When you integrate, going backwards, you're actually losing that 2!
So you we need to divide it out.

Answer 25

\[\Large\rm f(x)=\int\limits\limits 2\cos(2x)dx\]\[\Large\rm f(x)=2\int\limits\limits \cos(2x)dx\]\[\Large\rm f(x)=2\left(\frac{1}{2}\sin2x\right)+c\]Mmm what do you think of that shortcut? :O
Prefer to do the ole u-sub?

Answer 26

haha either way is fine but i feel like im probably more comfortable with u sub! :P

Answer 27

ya ya :) stick with what works i guess\[\Large\rm f(x)=\sin(2x)+C\]Ok so we've got this. I'm going to call it big C just so we can show that it's a different constant than before. It's not that same 0 necessarily.
Let's plug in our initial data to figure out our C value.

Answer 28

\(\Large\rm f(0)=0\)
x=0, y=0.
What's that give you for C? :3

Answer 29

0?

Answer 30

Uhhh ok good good good.\[\Large\rm f(x)=\sin(2x)\]

Answer 31

And then bam, plug in your pi/4, ya?

Answer 32

but where do i plug in the pie over 4?

Answer 33

You're evaluating the function f, at x=pi/4.
So you're replacing your x's with pi/4's

Answer 34

\[\Large\rm f(\color{orangered}{x})=\sin\left(2\color{orangered}{x}\right)\]Like this, ya? :)\[\Large\rm f\left(\color{orangered}{\frac{\pi}{4}}\right)=\sin\left(2\cdot\color{orangered}{\frac{\pi}{4}}\right)\]

Answer 35

okay! so should i keep my answer with a decimal?

Answer 36

Exact value is probably better.
Either method should get you the same value either way for this problem though,
because sin(pi/2) simplifies down really nicely.

Answer 37

so the answer is just 1?

Answer 38

Yay good job \c:/
f(pi/4) = 1

Answer 39

thank you so much!

Answer 40

Can we jump down to #12?
11 doesn't look very fun -_- maybe come back to that one lol

Answer 41

haha yeah thats fine!

Answer 42

\[\Large\rm g(x)=\int\limits_2^{3x} e^{t^4}dt\]To find the derivative of an integral, you're applying your Fundamental Theorem of Calculus, part 1.\[\Large\rm \frac{d}{dx}\int\limits_c^x f(t)dt=f(x)\]Does that look familiar maybe? :o

Answer 43

i want to say yes! its just been a while to be honest!

Answer 44

Lemme color things just to make sure this is making sense.\[\Large\rm g(x)=\int\limits\limits_2^{3x} \color{orangered}{e^{t^4}}dt\]
The idea is this...
you're starting with some messy function, the orange part, we'll call it g(t).
You assume some anti-derivative exists, call it G(t), but don't actually look for it.
Then you plug in your limits and do subtraction: G(3x)-G(2).
Then take derivative getting back to little g.

Answer 45

So you're integrating,
then you're differentiating,
getting back what you started with,
but in terms of a new variable.

Answer 46

We'll take our derivative to find g'(x),
\[\Large\rm g'(x)=\frac{d}{dx}\int\limits\limits\limits_2^{3x} \color{orangered}{e^{t^4}}dt\]

Answer 47

\[\Large\rm g'(x)=\frac{d}{dx}\int\limits\limits_2^{3x} \color{orangered}{g(t)}dt\]So if we generalize and call it something like g(t),
our first step is to integrate, giving us G(t).\[\Large\rm g'(x)=\frac{d}{dx}\left[G(t)|_2^{3x}\right]\]And then plug in our limits,\[\Large\rm g'(x)=\frac{d}{dx}\left[G(2x)-G(3)\right]\]And then finally apply the derivative.\[\Large\rm g'(x)=\color{orangered}{g(2x)}(2x)'-0\]The thing on the far right is just a constant since we plugged a 3 in, so it's derivative is 0.

Answer 48

What do you think, is this process too confusing? 0_o
I can simplify it down a little maybe.
I might be going into too much detail.

Answer 49

no i think it makes sense!

Answer 50

Woops, that was a 3x. not 2x's.\[\Large\rm g'(x)=\color{orangered}{g(3x)}(3x)'\]So what happens is, you end up with that same exponential that you started with,\[\Large\rm g'(x)=\color{orangered}{e^{(3x)^4}}(3x)'\]Except the t has been replaced by 3x.
And you end up with a little bit of something more because of the chain rule.
Your upper limit of integration was `more than simply x` so you end up with some stuff.
Looks like an extra 3 in this case, ya?\[\Large\rm g'(x)=\color{orangered}{e^{(3x)^4}}(3)\]

Answer 51

hahah yeah! that makes sense! itll take some practice but it does make sense!

Answer 52

#11 wants us to use the Limit/Reimann definition of integration to find this integral?
Mmm I'm a little rusty >.<

Lemme see if someone smarter can help out hehe
@dan815 @rational

Answer 53

thank you soo much for all your help!

Answer 54

np \c:/

Answer 55

for #11 see if this works
http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D%28%282%28i%2Fn%29%5E2%2B3%28i%2Fn%29-5%29*1%2Fn%29

Answer 56

@natasha.aries

Answer 57

thats the right riemann sum for "n" rectangles
you need to take the limit for the exact integral

Answer 58

for the -17n^2 etc..?

Answer 59

do you need to show work ?

Answer 60

yess i dooo

Answer 61

then you cannot let wolfram evaluate the partial sum
you need to work it using known formulas

Answer 62

\[\large \int\limits_0^1 (2x^2+3x-5)\,dx~~=~~ \lim\limits_{n\to\infty}\sum\limits_{i=1}^{n} \left(2(\frac{i}{n})^2+3(\frac{i}{n})-5\right)*\frac{1}{n}\]

Answer 63

the right side is the riemann sum, see if it makes sense and try to simplify if psble..

Answer 64

okay! thank u!!

Answer 65

hey we're not done yet,
we need to evaluate that partial sum and the limit

Answer 66

did u get how the riemann sum was set up ?

Answer 67

yeah sorry i actually have to go because im feeling light headed. im going to work on this problem in a couple of hours after i rest sorry

Answer 68

hope you can help me then but if not then thanks for the help!

Answer 69

sure tag me when you're back :)

Answer 70

thank you :)

Answer 71

im backk @rational

Answer 72

@rational can we continue this?