Author Topic: exam questions (Read 7333 times) Tweet Share

kdgamz · « **on:** September 29, 2009, 03:27:36 pm »

i cant seem to do this one (insight 2009)

Question

The length, X centimetres, of the wings of the type A butterfly has been found to have a probability density function :

f(x) = { 0.05e^(-0.05x) for x > 0
{ 0 for x ≤ 0

a) Find the mean length of the wings of the Type A butterfly.

thanks

Flaming_Arrow · « **Reply #1 on:** September 29, 2009, 03:36:03 pm »

evaluate $\int_0^{\infty} x* 0.05e^{-0.05x} dx = 20$

EvangelionZeta · « **Reply #2 on:** September 29, 2009, 03:38:07 pm »

Remember that for e^(-x), as x approaches infinity, the value of e^(-x) approaches approximately zero.

NE2000 · « **Reply #3 on:** September 29, 2009, 03:38:29 pm »

It's just a calc. question, use fnInt with 10000 as your upper bound. You need first year uni integration by parts to do that manually I think (correct me if I'm missing something here).

TrueTears · « **Reply #4 on:** September 29, 2009, 03:41:05 pm »

$x0.05e^{-0.05x}$

We require $\lim_{k \to \infty} \int_0^k x0.05e^{-0.05x} dx$

First consider $\frac{d}{dx} (xe^{-0.05x}) dx = (1 - 0.05x)e^{-0.05x} = e^{-0.05x} - 0.05xe^{-0.05x}$

Thus $\int 0.05xe^{-0.05x} dx = -(x+20)e^{-0.05x}$

Thus back to the integral at hand.

$\lim_{k \to \infty} \int_0^k x0.05e^{-0.05x} dx = \lim_{k \to \infty} [-(x+20)e^{-0.05x}]_0^k$

$= \lim_{k \to \infty} (-(k+20)e^{-0.05k}) - (-(0+20)e^{-0.05 \times 0})$

$= 0 - (-(0+20)e^{-0.05 \times 0})$

$= 20$

Notice the integral can also be integrated by parts.

NE2000 · « **Reply #5 on:** September 29, 2009, 03:48:12 pm »

lol this is methods TT...

Essentially TT integrated by recognition, but the knowledge of how you can integrate particular functions by diffing first is not for the VCE course so don't stress about it (it's good to know...but not to use in the exam)

TrueTears · « **Reply #6 on:** September 29, 2009, 03:49:44 pm »

lol as kamil would put it "Mathematical rigour ftw."

kurrymuncher · « **Reply #7 on:** September 29, 2009, 05:30:15 pm »

Quote from: NE2000 on September 29, 2009, 03:48:12 pm

lol this is methods TT...

Essentially TT integrated by recognition, but the knowledge of how you can integrate particular functions by diffing first is not for the VCE course so don't stress about it (it's good to know...but not to use in the exam)

So, its bad if we use integration by parts in the exam? will they even have questions in the exam that require integration by parts? I mean spesh lol

NE2000 · « **Reply #8 on:** September 29, 2009, 05:35:05 pm »

They shouldn't have questions that directly require you to use integration by parts. If you feel that parts is the way you ought to solve a question then just get out the calculator and solve it using that because that's what they want you to do.

kamil9876 · « **Reply #9 on:** September 29, 2009, 05:37:33 pm »

Quote from: TrueTears on September 29, 2009, 03:41:05 pm

$= \lim_{k \to \infty} (-(k+20)e^{-0.05k}) - (-(0+20)e^{-0.05 \times 0})$

$= 0 - (-(0+20)e^{-0.05 \times 0})$

That limit is not as trivial as you make it out to be. ( $\lim_{k \to \infty} ke^{-k}}$ ). I remember doing a similair question like this using this method in vce made me ask what is this limit indeed.

TrueTears · « **Reply #10 on:** September 29, 2009, 05:37:38 pm »

Quote from: kurrymuncher on September 29, 2009, 05:30:15 pm

Quote from: NE2000 on September 29, 2009, 03:48:12 pm
lol this is methods TT...

Essentially TT integrated by recognition, but the knowledge of how you can integrate particular functions by diffing first is not for the VCE course so don't stress about it (it's good to know...but not to use in the exam)

So, its bad if we use integration by parts in the exam? will they even have questions in the exam that require integration by parts? I mean spesh lol

Nah you won't lose any marks if you get the question right.

TrueTears · « **Reply #11 on:** September 29, 2009, 05:38:14 pm »

Quote from: kamil9876 on September 29, 2009, 05:37:33 pm

Quote from: TrueTears on September 29, 2009, 03:41:05 pm

$= \lim_{k \to \infty} (-(k+20)e^{-0.05k}) - (-(0+20)e^{-0.05 \times 0})$

$= 0 - (-(0+20)e^{-0.05 \times 0})$

That limit is not as trivial as you make it out to be. ( $\lim_{k \to \infty} ke^{-k}}$ ). I remember doing a similair question like this using this method in vce made me ask what is this limit indeed.

Indeed, but for VCE methods, that would suffice.

kamil9876 · « **Reply #12 on:** September 29, 2009, 05:47:53 pm »

What do you mean? You would never need to find $\lim_{k \to \infty} ke^{-k}}$ in vce methods anyway. How would you evaluate that in vce methods?

NE2000 · « **Reply #13 on:** September 29, 2009, 06:02:15 pm »

Quote from: kamil9876 on September 29, 2009, 05:47:53 pm

What do you mean? You would never need to find (Image removed from quote.) in vce methods anyway. How would you evaluate that in vce methods?

Do a quick sketch graph to show what you're saying (just suggesting what to do if such a situation arises)?

TrueTears · « **Reply #14 on:** September 29, 2009, 06:11:29 pm »

Quote from: kamil9876 on September 29, 2009, 05:47:53 pm

What do you mean? You would never need to find (Image removed from quote.) in vce methods anyway. How would you evaluate that in vce methods?

Yes you do.

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