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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on July 19, 2009, 01:59:08 pm

Title: PLease help
Post by: khalil on July 19, 2009, 01:59:08 pm: Question 12)c on page pg.481 in the essentials textbook!

The function f(x)= In(2x)

a) Find f^-1 and sketch the graphs of f and f^-1 on the one set of axes
b) Find the exact value of the area $\int_a^b f^-1(x) dx$ where b= In(4) and a=0
c) Find the exact value of $\int_a^b f(x) dx$ where b=2 and a=3/2

Im having trouble with c)
Title: Re: PLease help
Post by: Flaming_Arrow on July 19, 2009, 02:04:21 pm: it would be wise to post the question here so ppl who dont have access to the book can help you
Title: Re: PLease help
Post by: khalil on July 19, 2009, 05:34:38 pm: Im sure TT does :)
Title: Re: PLease help
Post by: dcc on July 19, 2009, 05:38:15 pm: The rest of us don't, however.
Title: Re: PLease help
Post by: cobby on July 19, 2009, 05:39:45 pm: Quote from: khalil on July 19, 2009, 01:59:08 pm
Question 12)c on page pg.481 in the essentials textbook!
why can't you just post the question?
Title: Re: PLease help
Post by: kat148 on July 19, 2009, 06:53:39 pm: there isn't a Question 12 on pg. 481...
Title: Re: PLease help
Post by: Flaming_Arrow on July 19, 2009, 06:54:43 pm: Quote from: kat148 on July 19, 2009, 06:53:39 pm
there isn't a Question 12 on pg. 481...

i think he's referring to the normal methods textbook(non-CAS)
Title: Re: PLease help
Post by: GerrySly on July 19, 2009, 06:58:52 pm: Haha, well this thread is just great
Title: Re: PLease help
Post by: dcc on July 19, 2009, 07:01:02 pm: About as great as enwiabe's ability to listen to other people's suggestions.
Title: Re: PLease help
Post by: kat148 on July 19, 2009, 07:03:55 pm: Quote from: Flaming_Arrow on July 19, 2009, 06:54:43 pm
Quote from: kat148 on July 19, 2009, 06:53:39 pm
there isn't a Question 12 on pg. 481...

i think he's referring to the normal methods textbook(non-CAS)
Hmm its not there in non-CAS either (assuming he's using the 4th Edition), but then it could be a typo *shrugs
Title: Re: PLease help
Post by: shinny on July 19, 2009, 07:07:02 pm: And this is why you type out the question OP! I've said before, if you want people to put the effort in to answer your question, put some effort into putting your question forward. In the end, it'll save you time anyway.
Title: Re: PLease help
Post by: kat148 on July 19, 2009, 07:25:44 pm: Was this your question Khali?

(from Chapter 14A of MM CAS textbook)

Q12) Cycling to school each morning, a girl goes through three sets of traffic lights. The probability of stopping at the first set is 0.6. If she stops at any one set, the probability that she has to stop at the next is 0.9. If she doesn’t have to stop at any one set, the probability that she doesn’t have to stop at the next is 0.7.

Use a tree diagram to find the probability that:

a) she has to stop at all three lights
b) she stops only at the second set of lights
c) she stops at exactly one set of lights
Title: Re: PLease help
Post by: khalil on July 19, 2009, 10:39:33 pm: I would be MORE than happy to type the question but i dont know how to display the integral sign.
This question is from the Ti-nspire CAS edition. In Exercise 12 J Q 12 c). It is the question about the anti derivative of log.
Thanks
Title: Re: PLease help
Post by: Flaming_Arrow on July 19, 2009, 10:41:32 pm: Quote from: khalil on July 19, 2009, 10:39:33 pm
I would be MORE than happy to type the question but i dont know how to display the integral sign.
This question is from the Ti-nspire CAS edition. In Exercise 12 J Q 12 c). It is the question about the anti derivative of log.
Thanks

$\int_a^b f(x) dx$

Code: [Select]
[tex] \int_a^b f(x) dx[/tex]
Title: Re: PLease help
Post by: khalil on July 19, 2009, 11:10:54 pm: The function f(x)= In(2x)

a) Find f^-1 and sketch the graphs of f and f^-1 on the one set of axes
b) Find the exact value of the area $\int_a^b f^-1(x) dx$ where b= In(4) and a=0
c) Find the exact value of $\int_a^b f(x) dx$ where b=2 and a=3/2

Im having trouble with c)
Title: Re: PLease help
Post by: khalil on July 19, 2009, 11:19:34 pm: What a great forum this turned out to be!
Title: Re: PLease help
Post by: cobby on July 20, 2009, 08:23:17 am: $\int_a^b f(x) dx = f(b) - f(a)$

$\int f(x) dx = [x*ln(x) + (ln(2)-1)*x]_\frac{3}{2}^2$

Then sub your upper limit $(b=2)$ and lower limit $(a=3/2)$ and subtract from there to get your final answer.

$[x*ln(x) + (ln(2)-1)]^2 - [x*ln(x) + (ln(2)-1)*x]_\frac{3}{2}$
Title: Re: PLease help
Post by: khalil on July 20, 2009, 09:16:08 am: Wait a second. How did you anti dervie log?
Title: Re: PLease help
Post by: GerrySly on July 20, 2009, 01:21:49 pm: $\begin{align*} y &= x\cdot \log_e{(x)}\\ \frac{dy}{dx} &= 1+\log_e{(x)}\\ \therefore \int \log_e{(x)} &= \int \frac{dy}{dx}-1.dx\\ &=y-x\\ &=x\cdot \log_e{(x)} - x \end{align*}$

Now using that knowledge

$\begin{align*} \int_\frac{3}{2}^2 \log_e{(x)}.dx&=\left [ x\cdot \log_e{(x)} - x \right ]_\frac{3}{2}^2\\ &=\left [ (2) \cdot \log_e{(2)} - (2) \right ] - \left [ \left ( \frac{3}{2} \right ) \cdot \log_e{\left ( \frac{3}{2} \right ) - \left ( \frac{3}{2} \right ) } \right ]\\ &=2\cdot \log_e{(2)} - 2 - \frac{3}{2} \cdot \log_e{\left ( \frac{3}{2} \right ) } + \frac{3}{2}\\ &=\frac{1}{2}\left [ \log_e{(16)} - \log_e{ \left ( \frac{27}{8} \right )} \right ] - \frac{1}{2}\\ &=\frac{1}{2}\log_e{ \left ( \frac{128}{27} \right ) } - \frac{1}{2}\\ &=\frac{1}{2}\left [ \log_e{ \left ( \frac{128}{27} \right ) } - 1 \right ]\\ \end{align*}$
Title: Re: PLease help
Post by: kamil9876 on July 20, 2009, 01:31:34 pm: Refer to attatched diagram

The way to do it without antidifferentiating log is to do it graphically. Notice how we can find the area we want by finding another area and then doing some subtraction etc. We can find the area by spltting the area in two, the rectangle at the bottom and the curvy part at the top. Finding the rectangle's area is easy, now all we need is a way to find the area of the curvy part at the top, and we can do that by integrating with respect to y:

$\int_c^d x dy$ This gives us the area trapped between the lines y=d, y=c, the y axis and the curve. To find that last curvy area we want we have to subtract the area we just found from the area of the big sideways rectangle:
$ Area_{curvy part}=2(d-c) - \int_c^d x dy$

Hence out total area is:
$ Area_{rectangle} + Area_{curvy part}= c(2 - \frac{3}{2}) + 2(d-c) - \int_c^d x dy$
Title: Re: PLease help
Post by: shinny on July 20, 2009, 04:54:46 pm: Since it wasn't set out as a integration by recognition question, the way kamil did it is what they were expecting you to do. I think they had a very similar question on a methods exam a few years back which involved doing the exact same thing.
Title: Re: PLease help
Post by: khalil on July 22, 2009, 06:38:51 pm: Thanks Kamil, but are you sure its possible to integrate along the y-axis?
Title: Re: PLease help
Post by: khalil on July 22, 2009, 06:40:12 pm: Also, Find $\int_0^3 f(3x) dx$ if $\int_0^9 f(x) dx$ = 5
Title: Re: PLease help
Post by: kamil9876 on July 22, 2009, 07:02:10 pm: Spec kids can just let u=3x here.

F(9) - F(0) = 5 where F is an antiderivative of f(x).

$\frac{d}{dx} F(3x)=3F'(3x)=3f(3x)$

$\therefore \int_0^3 3f(3x) dx=(F(3*3)-F(0))=(F(9)-F(0))=5$

Hence:

$\int_0^3 3f(3x) dx=5$
$\implies 3\int_0^3 f(3x) dx=5$

And the rest is trivial.
Title: Re: PLease help
Post by: kamil9876 on July 22, 2009, 07:05:49 pm: Quote from: khalil on July 22, 2009, 06:38:51 pm
Thanks Kamil, but are you sure its possible to integrate along the y-axis?

Yes, it's like looking at the graph sideways sort of. (area doesn't change). I.e: just treat the y-axis as the x-axis, it's arbitrary which is x and y.

Although I do like your question (if it inspires a rigorous Cauchy-like proof involving limits and Riemman sums).
Title: Re: PLease help
Post by: TrueTears on July 22, 2009, 07:16:53 pm: Quote from: khalil on July 22, 2009, 06:38:51 pm
Thanks Kamil, but are you sure its possible to integrate along the y-axis?
To integrate with respect to y is just using the inverse of your function which is in terms of x. It is in the methods course.
Title: Re: PLease help
Post by: khalil on July 22, 2009, 08:56:42 pm: Quote from: kamil9876 on July 22, 2009, 07:02:10 pm
Spec kids can just let u=3x here.

F(9) - F(0) = 5 where F is an antiderivative of f(x).

$\frac{d}{dx} F(3x)=3F'(3x)=3f(3x)$

$\therefore \int_0^3 3f(3x) dx=(F(3*3)-F(0))=(F(9)-F(0))=5$

Hence:

$\int_0^3 3f(3x) dx=5$
$\implies 3\int_0^3 f(3x) dx=5$

And the rest is trivial.

I dont understand...are you stating that both integrals =5?
Title: Re: PLease help
Post by: kamil9876 on July 22, 2009, 09:00:32 pm: which part specifically do you find unconvincing? Note the coefficient of 3 at the front, meaning the other integral is 5/3
Title: Re: PLease help
Post by: khalil on July 22, 2009, 09:49:45 pm: Could you explain it differently....I dont understand the method you have administered! Sorry
Title: Re: PLease help
Post by: kamil9876 on July 22, 2009, 10:16:05 pm: $\int_0^9 f(x) dx=F(9)-F(0)$ where F is an antiderivative of f(x).

using the value you are given for the first integral you can see:

$F(9)-F(0)=5$ (1)

Now notice that:

$\frac{d}{dx} (F(3x))=\frac{d(3x)}{dx} * \frac{F(3x)}{d(3x)}$ by the chain rule

$=3 * \frac{F(u)}{du}$ where u=3x

$=3f(u)$
$=3f(3x)$

Because $\frac{d}{dx}(F(3x))=3f(3x)$ we know that:

$\int_0^3 3f(3x) dx=[F(3x)]_0^3$
$=F(3*3)-F(0*3)$
$=F(9)-F(0)$ (2)

Subbing in equation (1) into (2) you get:

$\int_0^3 3f(3x) dx = 5$

But because of the wonderful property integrals that allows you to take out constant factors outside the integral you can change that equation by taking out the factor of 3:

$3\int_0^3 f(3x) dx = 5$

which gives $\int_0^3 f(3x)=\frac{5}{3}$
Title: Re: PLease help
Post by: khalil on July 23, 2009, 08:01:48 pm: Thanks Kamil, you really know you're stuff!!!!
Title: Re: PLease help
Post by: khalil on July 23, 2009, 08:05:03 pm: New Question:

Nathan knows that his probability of kicking more than four goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than four goals in the next game.
Title: Re: PLease help
Post by: kat148 on July 23, 2009, 08:12:15 pm: That same question was asked here
Title: Re: PLease help
Post by: khalil on July 23, 2009, 08:51:50 pm: Thanks Kat!
Title: Re: PLease help
Post by: khalil on July 30, 2009, 02:46:19 pm: A spinner is numbered from 0 to 5, and each of the six numbers has an equal chance of coming up. A player who bets $1 on any numbers wins $5 if that number comes up, otherwise the $1 is lost. What is the player's expected propit on the game?
Title: Re: PLease help
Post by: Mao on July 30, 2009, 10:43:10 pm: X=+$4, Pr = 1/6
X=-$1, Pr = 5/6

E(x) = 4/6 - 5/6 = -1/3