Question

Cindee, maf time! @Kayders1997

Answer 1

Yes :(

Answer 2

question? :D

Answer 3

Just 1 sec

Answer 4

I was helping someone with a question before you came we were almost done

Answer 5

k

Answer 6

Hopefully lol

Answer 7

It's a basic equation 7x-1=20

Answer 8

Okay the integral Sinpitdt

Answer 9

\[\large\rm \int\limits \sin(\pi t)~dt\]

Answer 10

How do you do that?

Answer 11

Magic.
Do you understand how to do this integral?\[\large\rm \int\limits \sin(u)~du\]

Answer 12

I barely understand u sub 😂

Answer 13

Well we're not dealing with that just yet.
Do you know how to find the anti-derivative of sin x?
That's all im asking so far.

Answer 14

Oh is it -cosx?

Answer 15

Ok great.
So if we were somehow able to transform our integral to sin(u),
we would know it turns into -cos(u).

Answer 16

Okay

Answer 17

\[\large\rm \int\limits\limits \sin(\color{orangered}{\pi t})~\color{royalblue}{dt}\]So we'll make the substitution \(\large\rm \color{orangered}{u=\pi t}\),
that will `partially` give us what we want,\[\large\rm \int\limits\limits\limits \sin(\color{orangered}{\pi t})~\color{royalblue}{dt}\quad=\quad \int\limits\limits\limits \sin(\color{orangered}{u})~\color{royalblue}{dt}\]

Answer 18

If we're going to do that,
we also must replace the differential dt with something involving du.

Answer 19

I don't understand what your supposed to choose at u, the other part that we didn't antiderivative at the beginning?

Answer 20

In general, you're looking for a \(\large\rm u\) and \(\large\rm u'\),
something and it's derivative.

It's a little more simple than that in this case though.
We have sin(pi t) but we know how to integrate sin(x) or sin(u),
so we're replacing the (pi t) by u.

Answer 21

We're using our understanding of the basic integral to simplify it down,
that's how we know what to choose,
whatever u makes the problem easier.

Answer 22

It won't always be immediately obvious though.

Answer 23

Okay I'm understanding more now

Answer 24

We also need to replace the differential with something involving u.
So start here with your substitution,\[\large\rm u=\pi t\]and take a derivative, with respect to t.

Answer 25

Okay so du/dt=pi?

Answer 26

Ok great.
Here is another way we can write this,\[\large\rm \frac{du}{dt}=\pi\qquad\qquad\to\qquad\qquad du=\pi~dt\]You can think of this process as "multiplying" the dt to the other side.
That's not entirely accurate since dt is an infinitesimal quantity,
but it follows the same rule.

Answer 27

Okay that makes sense

Answer 28

Yeah I was like ugghhhhhh idk lol

Answer 29

Notice that this dt is the same dt in our integral,\[\large\rm du=\pi~\color{royalblue}{dt}\]So think of it as a veriable that you would like to isolate.
I guess we can divide the pi to the other side, ya?

Answer 30

Yeah

Answer 31

This is getting really messy

Answer 32

\[\large\rm \color{royalblue}{\frac{1}{\pi}du=dt}\]

Answer 33

Eh it's not bad :D

Answer 34

Let's group up our information so we don't lose anything.
One sec,

Answer 35

Okay, and none relevant question is this the last harder concept in calculus?

Answer 36

Hardest

Answer 37

No, power series is much much harder (at least for me).
But I tend to think of these types of integrals as really easy,
so maybe it's just different for different people.

Answer 38

Omg :O I don't like these but someone new has been teaching for about 1 week now

Answer 39

We start with this integral,\[\large\rm \int\limits\limits\limits \sin(\color{orangered}{\pi t})~\color{royalblue}{dt}\]and we determined that this substitution would be appropriate:
\[\large\rm \color{orangered}{u=\pi t}\]\[\large\rm \color{royalblue}{\frac{1}{\pi}du=dt}\]Making these replacements gives us,\[\large\rm \int\limits\limits\limits\limits \sin(\color{orangered}{u})~\color{royalblue}{\frac{1}{\pi}du}\]

Answer 40

I hope the color is helping a little bit.
Pay attention to that.

I'm just replacing orange with orange,
and blue with blue.

Answer 41

Yes the color helps a lot thank you for that

Answer 42

From this point, let's pull the 1/pi out in front of the integral,\[\large\rm \frac{1}{\pi}\int\limits \sin(u)~du\]

Answer 43

Because it's a constant!

Answer 44

yes, constant coefficient.

Answer 45

And now, integrate.

Answer 46

So -cosu du/dt?

Answer 47

0_o

Answer 48

Don't know about the second part :O

Answer 49

what second part?

Answer 50

Well the du/dt

Answer 51

there is no du/dt in the integral,
i'm not sure what you're talking about :U

Answer 52

Idk either I just got confused when you wanted me to anti differentiate

Answer 53

Well you told me earlier that the anti-derivative of sin(x) is -cos(x), right?
So,\[\large\rm \frac{1}{\pi}\color{purple}{\int\limits\limits \sin(u)~du}\qquad=\quad \frac{1}{\pi}\color{purple}{(-\cos(u))}\]

Answer 54

What happened to du 0_o

Answer 55

The du and the big squiggly bar are buddies.
They come and go as a package.

When you perform the "integration" step,
they both disappear.
Not just the squiggly bar, but also the differential.

Answer 56

Oh weird

Answer 57

So is that the answer?

Answer 58

\[\large\rm \int\limits x~dx=\frac12x^2\]Both the bar and the dx disspear

Answer 59

example ^

Answer 60

Well we have one final step,\[\large\rm -\frac{1}{\pi}\cos(\color{orangered}{u})\]Remember that the problem was given to us in terms of \(\large\rm t\), not this silly \(\large\rm u\),
so finally, we need to `undo our subistution`.

Answer 61

Ohhhhhhhh

Answer 62

So -1/pi cos(pi(t))?

Answer 63

\[\large\rm \int\limits \sin(\pi t)~dt\quad=\quad -\frac{1}{\pi}\cos(\pi t)+C\]Good good good.

Answer 64

Oh yeah that darn +c

Answer 65

There is an easier, more intuitive way to do these,
but you're just starting to learn u-sub,
so we shouldn't try to stuff too much into your brain right now.

Answer 66

Okay plus if I SOMEHOW lol learned it I would probably want to do your way when it says on an exam solve by u substitution

Answer 67

lol ya that'd be a problem XD

Answer 68

Are you tired or could we do one of the other problems the parts

Answer 69

Ooo parts? bring it.

Answer 70

Okay so intergral of tsin2tdt

Answer 71

I know you need to pick out a u v dv and du

Answer 72

You need to pick out \(\large\rm u\) and \(\large\rm dv\).

From there, you'll `differentiate` to get a \(\large\rm du\),
and you'll `integrate` to get a \(\large\rm u\).

So realize...

Answer 73

So realize?

Answer 74

that whatever you pick for your \(\large\rm u\) is going to be differentiated.
Try to think about the differentiation process...
Generally speaking, it makes things simpler, it breaks them down.

At least this is true with polynomials right,
if we have t^2, it turns into 2t, then 2, then 0 through multiple differentiations.

Answer 75

Right

Answer 76

So the easiest thing to differeniate here is t

Answer 77

But sine does work so nicely,
sine, cosine, -sine, -cosine, sine
it's very circular, not nice for differentiation.
Doesn't break down like we want.

Answer 78

Good good good, so the \(\large\rm t\) is what we want for our \(\large\rm u\)

Answer 79

Which means everything else is our \(\large\rm dv\).

Answer 80

\[\large\rm \int\limits t \sin(2t)dt\]Oh they gave us a stinker...
I guess we do have to learn this shortcut trick :(

You really really don't want to apply a u-substitution within a parts problem.
It get's really burdensome.

Answer 81

Okay, yeah they never taught us really

Answer 82

Do you remember how to take this derivative?\[\large\rm \left(e^{2x}\right)'\]

Answer 83

I think so :)

Answer 84

Isn't it... 2e^2x?

Answer 85

Good, you end up `multiplying` by that extra 2,
because of the chain rule.

Answer 86

So what would happen here?\[\large\rm \int\limits 2e^{2x}~dx\]

Answer 87

Why did we change it to e though?

Answer 88

We'll relate it to the original problem in a moment :U
You have to learn this trick though!

Answer 89

Okay!

Answer 90

Ummm I'm not sure :(

Answer 91

Well, if the derivative of e^(2x) is 2e^(2x),
then the anti-derivative of 2e^(2x) is e^(2x)
right?

Answer 92

Lol omg yes

Answer 93

\[\large\rm \int\limits\limits 2e^{2x}~dx\quad=\quad e^{2x}\]

Answer 94

But what happened to the 2 in this process?
Try to think about mathematically...
which simple operation got rid of the 2?

Answer 95

2/2 o_O

Answer 96

\[\large\rm \int\limits\limits\limits 2e^{2x}~dx\quad=\quad \frac22e^{2x}\quad=\quad e^{2x}\]Good good good,
we actually had to `divide` by the 2 when we integrated, right?

Answer 97

Yes

Answer 98

So what about this integral?\[\large\rm \int\limits e^{2x}dx\]Any ideas? :)