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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: jack_chay on September 12, 2013, 08:14:32 pm

Title: Final Exam HELP
Post by: jack_chay on September 12, 2013, 08:14:32 pm: I have some questions that my teacher has given me in practise exams.

integral of 5 to 1 ( f(x) ) dx = 6

then

2 times integral of 5 to 1 ( f(x) + 3) dx equals:

thanks in advance
Title: Re: Final Exam HELP
Post by: Alwin on September 12, 2013, 08:16:49 pm: Quote from: jack_chay on September 12, 2013, 08:14:32 pm
I have some questions that my teacher has given me in practise exams.

integral of 5 to 1 ( f(x) ) dx = 6

then

2 times integral of 5 to 1 ( f(x) + 3) dx equals:

thanks in advance

Consider f(x) = 3/2 :)

Try it and see what you get!
Title: Re: Final Exam HELP
Post by: rhinwarr on September 12, 2013, 08:23:22 pm: Use the rule that:
Integral of f(x) + c is equal to the integral of f(x) + integral of c

Therefore:
2 times integral of 5 to 1 ( f(x) + 3) dx equals:
2 times integral of 5 to 1 f(x) dx + integral of 5 to 1 (3) dx
=2 times 6 times (15 - 3)
=12 times 12
=144
Title: Re: Final Exam HELP
Post by: jack_chay on September 12, 2013, 08:26:59 pm: um... but the only options are 9, 15, 18, 24 and 36, there are no 144??
Title: Re: Final Exam HELP
Post by: jack_chay on September 12, 2013, 08:28:44 pm: don't worry I got it, it's 36, thank you :)
Title: Re: Final Exam HELP
Post by: rhinwarr on September 12, 2013, 08:31:22 pm: Woops, I times it instead of added it.
2 x (6 + 12) = 36 :)
Title: Re: Final Exam HELP
Post by: Homer on September 12, 2013, 09:24:46 pm: was this from MAV 2012? if yes then it is 36
Title: Re: Final Exam HELP
Post by: jack_chay on September 18, 2013, 09:14:17 pm: all the jibberish...

i have no idea what the question is asking...

thank you in advance
Title: Re: Final Exam HELP
Post by: SocialRhubarb on September 18, 2013, 09:22:22 pm: Haha, where's this from?

I'm pretty sure it's just asking for the integral of $x^2$ from 0 to 4.
Title: Re: Final Exam HELP
Post by: b^3 on September 18, 2013, 09:38:15 pm: What the question is saying is that we're taking the function $x^{2}$ and over the interval $0\leq x\leq4$ , we're going to divide the area under this curve up into rectangles, in this case $n$ rectangles. The width of a rectangle here will be $\delta x$ , so you can interpret it as a 'small change' in $x$ , or that the width of the rectangles is a small portion of our interval. Now we're going to look at what happens when we sum the area of these rectangles, but make the rectangle width go to zero. That is as $\delta x\rightarrow 0$ . What we end up with is just the area under the curve in that interval, which as SocialRhubarb has said, will just be the definite integral of the function over that interval.

Hope that makes sense.
Title: Re: Final Exam HELP
Post by: Phy124 on September 18, 2013, 09:39:55 pm: Ok so remember when you first learnt about integration you did it by making rectangles under the graph and finding the area of these rectangles?

Well when you did that, maybe you had 3 rectangles or 4 or 5 or 6 etc.

This question is stating that we are making $n$ rectangles and summing the area of these rectangles (with width $\delta x$ i.e. the right value of $x$ of the rectangle , $x_i$ , minus the left value of $x$ of the rectangle, $x_{i-1}$ , and height $x_i^2$ i.e. the value of $x$ to the right squared).

The $n$ value is practically infinite as we are finding the sum at which the width of the rectangles is approaching zero.

This is the underlying principle which integration is based on but not many people seem to know much about. You'd be much more familiar with the form $\int^4_0 x^2 dx$ , no?
Title: Re: Final Exam HELP
Post by: jack_chay on September 18, 2013, 10:13:21 pm: how do you know it is o to 4, i thought it was 1 to 4...
Title: Re: Final Exam HELP
Post by: Phy124 on September 18, 2013, 10:17:36 pm: The key parts to that are;

Quote
$\text{The interval}\ [0,4]$ ...

... $\text{where} \ 0 = x_0$ ...

So it's saying you're doing it over the interval [0,4] and if you don't pick it up from that the second part I quoted is saying that $x_0$ , the first point you're taking (that is, the left value of the left most rectangle) is $x=0$ :)
Title: Re: Final Exam HELP
Post by: jack_chay on September 19, 2013, 08:04:08 am: oh yeah.. hehehe sorry thank you ;D