Question

average value, ftc, and differential equation

Answer 1

@TuringTest @dpaInc

Answer 2

Wow, how far have you gotten on your own so far?

Answer 3

Well I am having trouble on the first part

Answer 4

If the average value of a function \(y=f(x)\) on the interval \([a,b]\) is \(\bar{y}\), then \(\displaystyle\int_a^b f(x)\,\text{d}x=(b-a)\bar{y}\). Since the average value of \(f\) over the interval \([0,k]\) is \(\displaystyle\frac{f(0)+f(k)}{2}\), \(\displaystyle\int_0^k f(x)\,\text{d}x=k\frac{f(0)+f(k)}{2}\).

Answer 5

btw, those are definitions...

Answer 6

why do we multiply k in the last part of your response

Answer 7

I think because the integral is (interval length)*average.

Answer 8

\(k-0=k\)...

Answer 9

oh okay.

Answer 10

so -7.5?

Answer 11

sorry -30

Answer 12

sorry... that image doesn't want to seem download to my computer...:(

Answer 13

I would think -4, if the average on the interval is -1

Answer 14

Yes, I got -4 as well...

Do you know how to do part B?

Answer 15

Wait why is it -4 shouldnt it be 4* f(0) + f(4) /2

Answer 16

2nd part: well, that is the average for f(x) up to the point. I'm guessing that has something to do with it.

Answer 17

4*(f(0)+f(4))/2=2*(f(0)+f(4))=2*(3-5)=2*-2=-4

Answer 18

Danielle, remember parenthesization 4*(f(0)+f(4))/2

Answer 19

Oh yeah. I multiplied instead of adding them.

Answer 20

It tells you to use FTC so it's already pretty easy...

For the last part we had already derived the formula \[\int_0^x f(t)\,\text{d}t=x\frac{f(0)+3}{2}=\frac{1}{2}xf(x)+\frac{3}{2}x,\]So now we simply differentiate both sides with respect to x:
\[\frac{\text{d}}{\text{d}x}\int_0^xf(t)\,\text{d}t=f(x)~~~~(\text{by the FTC})\]\[\frac{\text{d}}{\text{d}x}\int_0^xf(t)\,\text{d}t=f(x)=\frac{\text{d}}{\text{d}x}\Big[\frac{1}{2}xf(x)+\frac{3}{2}x\Big]=\frac{1}{2}f(x)+\frac{1}{2}xf'(x)+\frac{3}{2}\]Product rule, of course, was used here.

Answer 21

So now we have\[f(x)=\frac{1}{2}f(x)+\frac{1}{2}xf'(x)+\frac{3}{2}\] and we need to solve for f'(x).

Answer 22

Questions?

Answer 23

I have a question about part b.

Answer 24

What?

Answer 25

How would i use ftc to prove that..?

Answer 26

The part where I said "by the FTC" is the part where you use FTC. :)

Answer 27

I thought the ftc was like integral of a to b of f(x) = f(b) - f(a)

Answer 28

There are actually two different "Fundamental Theorems of Calculus", the first one, caled the "First Fundamental Theorem of Calculus" was the one you mentioned. The second one, the "Second Fundamental Theorem of Calculus" was the one I used.

Tricky, huh?

Answer 29

I just read about the second one. so whenever its from 0 you don't include that part right?

Answer 30

Actually, the lower bound of the integral could be \(any~number~whatsoever\). The theorem says:\[\frac{d}{dx}\int_a^x f(t)\,dt=f(x)\]For \(any\) number \(a\).

Answer 31

That's how they try to trick you on the AP Calculus AB exam :)

Answer 32

Tricky. Tricky

Answer 33

I actually have to go... Maybe you could get someone else for C; once you get there.

Answer 34

Could you just start me off on that one? I just finished B

Answer 35

@helder_edwin can you help me with part c?

Answer 36

actually i just figured it out.