Question

@jim_thompson5910

Answer 1

So I was given 3 seprate packet things if you will, the first being small and simple. It is on the average value of a function and the 2nd fundamental theorem. I'm fine with average value, its the 2nd fund. I need a refreshment one

Answer 2

ok we'll see how much we can get through

Answer 3

Coolio we can do more tomorrow too, maybe we can set a time or whatever idk

Answer 4

that works

Answer 5

Working on pulling up a picture now

Answer 6

ok

Answer 7

Wait even better

http://www.mastermathmentor.com/mmm/Free.aspx?bin=calc.AB%20Manual&file=AB32%20avg%202nd%20FTC.pdf

Answer 8

Starting at Pg 2 right after example 2

Answer 9

ok one sec

Answer 10

The fundamental theorem of calculus (FTC) comes in two parts. Some books state each part differently in different orders, but the ideas are still the same


Part 1)

if \(\Large y = \int_{a}^{x} g(t)dt\), then \(\Large \frac{dy}{dx} = g(x)\). This shows us the connection between derivatives and integrals. They are opposite operations (more or less)

-------------------------------------------------------

Part 2)

\[\Large \int_{a}^{b} f'(x)dx = f(b)-f(a)\]

this second part allows us to compute areas under curves between two endpoints. Notice how I have f ' (x) in the integral. The integral undoes the derivative to get back to f(x). We don't have to worry about the +C when it comes to definite integrals

Answer 11

hopefully those two parts are familiar?

Answer 12

Yes I understand the fundamental theorem and all that very well

Answer 13

I know the 2nd has something to do with plugging xs in for ts or something like that

Answer 14

so under example 2, do you see how to compute that integral?

Answer 15

I think

x^2-4x+1?

Answer 16

that's what would happen when you take the derivative of the integral

Answer 17

Oh okay

1 to x of t^3/3-2t^2+t?

Answer 18

yes, that is the integral, with +C added to the end

Answer 19

well no, slight typo

Answer 20

it should be this
\[\Large \int (t^2 - 4t + 1)dt = \frac{t^3}{3} - 2t^2 + t + C\]

Answer 21

Thats what I put except the +c

Answer 22

when you add in the limits of integration, you get

\[\large \int_{1}^{x} (t^2 - 4t + 1)dt = \left[\frac{t^3}{3} - 2t^2 + t + C\right]_{1}^{x}\]

\[\large \int_{1}^{x} (t^2 - 4t + 1)dt = \left[\frac{x^3}{3} - 2x^2 + x + C\right]- \left[\frac{1^3}{3} - 2(1)^2 + 1 + C\right]\]

\[\large \int_{1}^{x} (t^2 - 4t + 1)dt = \frac{x^3}{3} - 2x^2 + x + C- \frac{1}{3} + 2 - 1 - C\]

\[\large \int_{1}^{x} (t^2 - 4t + 1)dt = \frac{x^3}{3} - 2x^2 + x + \frac{2}{3}\]

Answer 23

with definite integrals, those "C"s will subtract off which is why you can ignore the +C with definite integrals

Answer 24

now when you take the derivative of both sides of

\[\large \int_{1}^{x} (t^2 - 4t + 1)dt = \frac{x^3}{3} - 2x^2 + x + \frac{2}{3}\]

you will get

\[\large \frac{d}{dx}\int_{1}^{x} (t^2 - 4t + 1)dt = \frac{d}{dx}(\frac{x^3}{3} - 2x^2 + x + \frac{2}{3})\]

\[\large \frac{d}{dx}\int_{1}^{x} (t^2 - 4t + 1)dt = x^2 - 4x + 1\]

Answer 25

so that proves that the derivative of the integral is as simple as replacing the 't' with 'x'

Answer 26

Oh okay

Answer 27

it's not always this simple since the chain rule may come into play, but right now we don't have to worry about the chain rule

Answer 28

Yeah thats why i knew I needed help lol

Answer 29

so the first one, a

Answer 30

the key is to look at the limits of integration

if you have it in the form

\[\Large \int_{a}^{x}\]

where 'a' is a fixed constant and x is the variable, then it's as simple as replacing 't' with x. That's it

Answer 31

however, if the 'x' is more complicated, like 'x^2' in part c, then you need to use the chain rule

Answer 32

Yeah and if x is on the bottom, flip it and put a - in front of the integral

Answer 33

correct

Answer 34

because \[\Large \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx\]

Answer 35

So then a is just


\[\sqrt{x^2-1}\]

Answer 36

yes it is

Answer 37

Is there any like math way of proving it/showing it? I have to actully do the integral stuff right?

Answer 38

Since the 2nd part will just cancel because 1-1 lol

Answer 39

no, you don't have to actually compute the integral because you have the 2nd fundamental theorem of calculus (in the box above part a)

Answer 40

Oh cool

Answer 41

\[\Large \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right] = f(x)\]

Answer 42

b is just -xsinx?

Answer 43

and you stated a handy trick, which is

\[\Large \frac{d}{dx}\left[\int_{x}^{a}f(t)dt\right] = \frac{d}{dx}\left[-\int_{a}^{x}f(t)dt\right] = -f(x)\]

Answer 44

yep, b is correct

Answer 45

Okay so now c, ew lol

Answer 46

You do, I watch and learn, then I know from there on how to do lol

Answer 47

Usually I'm good once you show me 1 problem on it

Answer 48

ok one sec on while I think on a way to approach this one

Answer 49

Okie dokie

Answer 50

actually, nvm, I'm thinking of something more complicated that isn't needed here

anyways, we'll use the other piece of what they give you in the box

\[\Large \frac{d}{dx}\left[\int_{a}^{u}f(t)dt\right] = f(u)\frac{du}{dx}\]

where u is a function of x

Answer 51

Ohhh I think I see
So you do the same thing but you have to multiply it by du?

Answer 52

2xcosx^2

Answer 53

part c)

f(t) = cos(t)
a = pi/2
u = x^2
du/dx = 2x

this means

\[\Large \frac{d}{dx}\left[\int_{a}^{u}f(t)dt\right] = f(u)\frac{du}{dx}\]

\[\Large \frac{d}{dx}\left[\int_{\pi/2}^{x^2}\cos(t)dt\right] = \cos(x^2)*2x\]

\[\Large \frac{d}{dx}\left[\int_{\pi/2}^{x^2}\cos(t)dt\right] = 2x\cos(x^2)\]

Answer 54

yep that's correct

Answer 55

Cool go me
I knew it was easy somehow, just needed a refresher lol

Answer 56

yeah it's probably the notation that's a bit strange at first, but after a while it's relatively simple

Answer 57

I did all of 1 and 2

Answer 58

alright

Answer 59

And 3 I did that

Can we do 4 and 5?

Answer 60

t = 0 represents 9 AM
t = 12 represents 9 PM (12 hrs later)

Answer 61

say you had a function f(x)

let M = average value of f(x) on the interval from x = a to x = b

to find the value of M, we use this formula

\[\Large M = \frac{1}{b-a}\int_{a}^{b}f(x)dx\]

Answer 62

Yeah that I know

Answer 63

Just not 100% sure on how to integrate the 14sin(pit/12) lol

Answer 64

In this case, a = 0, b = 12, f(x) is really the function T(t) = 50+14sin(pi*t/12)

which means you need to compute

\[\Large M = \frac{1}{12-0}\int_{0}^{12}(50+14\sin(\pi*t/12))dt\]

Answer 65

have you learned about u-substitution?

Answer 66

Yeha
I wrote that down already

Answer 67

Yes I'm good at it but the break has kinda like.. made it gone away haha

Answer 68

ok so I would let

u = pi*t/12

derive both sides with respect to t to get

du/dt = pi/12

then solve for dt

12du = pi*dt

dt = (12du)/pi