Question

calculus help

Answer 1

help

Answer 2

First, take the integral of \(\sf 4x^3 \Rightarrow \color{red}{x^4}\) but since this is a definite integral, you must evaluate this from the given bounds, 1<x<2

So: \(\sf \color{blue}{(2)^4-(1)^4}\)
and you can do the math from here.

Answer 3

@abb0t 15?

Answer 4

Yes.

Answer 5

and then from there what do i do to get C?

Answer 6

@ganeshie8 help

Answer 7

\[\Large\rm \int\limits_1^2 4x^3~dx=x^4|_1^2=2^4-1^4\]But they're telling us that this integral is also equal to:\[\Large\rm \int\limits_1^2 4x^3~dx=4c^3(2-1)\]right?

Answer 8

So to find our c, I think we can just set these equal to one another and do some math stuff from there.\[\Large\rm 2^4-1^4=4c^3(2-1)\]

Answer 9

i get 1.55, what about you?

Answer 10

@zepdrix

Answer 11

Hmm yah that sounds about right :)

Answer 12

im getting it wrong

Answer 13

Hmmmm do they want an exact answer maybe? :o
\(\Large\rm \sqrt[3]{4}\)
Try that. If it doesn't work, maybe we go back to the drawing board.
Or put it in as 4^(1/3) if you don't have special characters.

Answer 14

Oh it's not 4^(1/3).. derp I'm dumb >.<
It was 15/4 to that power, yes?

Answer 15

Where you at prote? >.<
You remember that guy prote from that movie k-pax? yah that was awesome :U

Answer 16

wait what?

Answer 17

(15/4)^(1/3)?

Answer 18

Yah does that maybe work? :o

Answer 19

nope :(

Answer 20

Hmm ok lemme look at the problem again. Maybe I did something silly...

Answer 21

its weird that in the book it says on the interval [-1,2]

Answer 22

oh... then you'd you put [1,2]? lol typo? :D

Answer 23

nvm

Answer 24

its on 1,2

Answer 25

not -1,2

Answer 26

Hmm I'm running out of ideas :c
Do you get unlimited guesses?
Wanna try punching in the answer for the interval [-1,2] maybe and see if they messed something up?

What does the website say? Does it say to round to any specific number of decimals?

Answer 27

yeah

Answer 28

I asked like 4 questions.. yeah to what? 0_o lol

Answer 29

unlimited guesses

Answer 30

k

Answer 31

heres the problem

Answer 32

oh that's the whole thing? :O
cool.

Hmm im running out of ideas :c
i guess try more decimals? 1.554, or 1.5536

Answer 33

thanks it worked this time

Answer 34

yay team!

Answer 35

heres another problem i need help on

Answer 36

The average value of a function is given by:\[\Large\rm f_{ave}=\frac{1}{b-a}\int\limits_a^b f(x)dx\]So for our problem we can find the average by inputting the data,\[\Large\rm L_{ave}=\frac{1}{b-a}\int\limits\limits_a^b L(t)dt\]

Answer 37

ok

Answer 38

\[\Large\rm L_{ave}=\frac{1}{20-0}\int\limits\limits\limits\limits_0^{20} t e^{-.01t^2}dt\]

Answer 39

Oh oh they want the average over the `first three years`, my bad.
Didn't read that closely enough.

Answer 40

\[\Large\rm L_{(\text{ave over first 3 years})}=\frac{1}{3-0}\int\limits_0^{3} t e^{-.01t^2}dt\]Mmmmm so something like that, yah?

Answer 41

A u-sub will take care of that integral nicely.

Answer 42

\[\Large\rm \frac{1}{3}\int\limits_0^3 e^{-.01t^2}\left(t~dt\right)\]
\[\Large\rm u=-.01t^2, \qquad\qquad du=?\]

Answer 43

im doing that right?

Answer 44

lol yes :D it's your problem :3

Answer 45

ohh im on it

Answer 46

im confused i get this

Answer 47

\[\Large\rm u=-.01t^2, \qquad\qquad du=-.02t~dt\qquad\to\qquad -50du=t~dt\]Plugging in gives us:\[\Large\rm \frac{1}{3}\int\limits_{t=0}^{t=3} e^{u}\left(-50~du\right)\]

Answer 48

Integrating gives,\[\Large\rm -\frac{50}{3}e^u=-\frac{50}{3}e^{-.01t^2}|_0^3\]

Answer 49

\[\Large\rm -\frac{50}{3}\left[e^{-.09}-e^0\right]\]Distributing the negative gives,\[\Large\rm \frac{50}{3}\left[1-e^{-.09}\right]\]

Answer 50

Which is almost exactly what you came up with.
I just cant figure out where the decimal on the 3 is coming from.

Answer 51

Oh and you missed a negative sign somewhere I think. Maybe in your du.

Answer 52

ok so what would "part a" be?

Answer 53

Umm I dunno, toss this into a calculator or something.
\[\Large\rm \frac{50}{3}\left[1-e^{-.09}\right]\]I'm pretty sure these calculations are correct up to this point.
Use three decimal points maybe, that seemed to work out for us last time.

Answer 54

ok im doing that right now

Answer 55

got 1.4344

Answer 56

what about part b?

Answer 57

\[\Large\rm \frac{1}{20-0}\int\limits_0^{20}f(t)~dt=f(t_1)\]So this average will equal the function at some specific time \(\Large\rm t_1\).
Hmmm thinking...

Answer 58

the reason why i ask you to do it is because im preparing a study for me to study for my final exam on august 8th

Answer 59

study guide*

Answer 60

I think the phrase "the average level of carbon dioxide actually occurs" in part b) refers to the average level found in part a).

Answer 61

Hmm I'm a little stumped in part b.
I don't think there is a way to do it algebraically.
Integrating eventually gives...\[\Large\rm \frac{5}{2}\left[1-e^{-.2}\right]\approx .453\]And we want to know what value of t makes the function equivalent to this value.\[\Large\rm t e^{-.01t^2}=.453\]

Answer 62

I must've set it up wrong, this doesn't have a nice solution :OC

Answer 63

@aum any ideas? D':

Answer 64

I think the phrase "the average level of carbon dioxide actually occurs" in part b) refers to the average level found in part a).

Use a graphing calculator to plot \(y = te^{-0.01t^2}\) and \(y = 1.4345\) and find
the points of intersection, t1 and t2. Round t1 and t2 to two decimal places.

I am getting t1 = 1.47 and t2 = 15.41

Answer 65

@aum got them right thanks

Answer 66

You are welcome.

Answer 67

cool c:

Answer 68

one more question and im done guys

Answer 69

\[
\Large\rm f_{ave}=\frac{1}{b-a}\int\limits_a^b f(x)dx =
\frac{1}{\pi/2-0}\int\limits_0^{\pi/2} (2\sin(x)-\cos(x))dx
\]

Answer 70

ok @aum