• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Intergration (2 Viewers)

Thecorey0

Member
Joined
Oct 28, 2008
Messages
428
Location
Goldstein
Gender
Male
HSC
2009
I am assuming the function is x^(2/3). If so there is no solution, because it is not defined for x<0. Therefore you can not evaluate the integral from 8 to -8. That is unless I am missing something.
 

Casper_18

Member
Joined
Mar 4, 2009
Messages
39
Gender
Male
HSC
2009


How is the value of -8^(5/3)/5 found?
when you do this you have to consider the graph. Seeing as though they are not asking for an area and what you are integrating is an even function. i.e. F(x) = F(-x) then it becomes:



so in your case



Hope this helps you
 

kaz1

et tu
Joined
Mar 6, 2007
Messages
6,960
Location
Vespucci Beach
Gender
Undisclosed
HSC
2009
Uni Grad
2018
I am assuming the function is x^(2/3). If so there is no solution, because it is not defined for x<0. Therefore you can not evaluate the integral from 8 to -8. That is unless I am missing something.
You can find the cube root of a negative number.
 

Casper_18

Member
Joined
Mar 4, 2009
Messages
39
Gender
Male
HSC
2009
You can find the cube root of a negative number.
Correct and if you remember your index laws anyway x^(2/3) you are squaring and then taking the cube root so you are taking a positive cube root anyway. because any number squared is positive in the real number system.
 

Aerath

Retired
Joined
May 10, 2007
Messages
10,169
Gender
Undisclosed
HSC
N/A
Why isn't x defined for x <0?

For instance, y = x^2. Does this mean that x cannot be -1?

[I think you mean y cannot be less than 0]
 

Thecorey0

Member
Joined
Oct 28, 2008
Messages
428
Location
Goldstein
Gender
Male
HSC
2009
Why isn't x defined for x <0?

For instance, y = x^2. Does this mean that x cannot be -1?

[I think you mean y cannot be less than 0]
I'm sorry, my mistake. I attempted the question and did not get the answer the poster asked for. So I entered the function on a graphing calculator, and simply observed that it only existed in the upper right quadrant.
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
It does seem you can take the cube root of a negative... But my calculator somehow, when I entered the -8^(5/3) gave me an error...

Any1 have any ideas about that? Is it because im using a sharp?
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Do you know why that happens? Is the casio calculators better?
 

Iruka

Member
Joined
Jan 25, 2006
Messages
544
Gender
Undisclosed
HSC
N/A
I believe some calculators find the cube root (or any other root) via an algorithm that takes logarithms first. That is why they are giving you a math error - you can't take the log of a negative number.
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Thanks Iruka.

It just that I am worried, that using a sharp calculator will give me a slight disadvantage or something. I know such novelties shouldnt even be considered at my level but...
 

Casper_18

Member
Joined
Mar 4, 2009
Messages
39
Gender
Male
HSC
2009
I have used the SHARP calculator since year 7, the exact same one. I dont Get math errors when I type in -8^(5/3)/5 though. The sharp calcuator is a very good calculator. Although I have found that Casio are better at calculating fractions of larger decimals.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top