| Blogs > Raithed |
|
Raithed
China7078 Posts
sketch the region enclosed by the curves and find the enclosed area a. y = x^3; y = x; x = 0; x = 1/2 b. y = x^2 + 4; y = 6-x c. x^3 - 2x^2; y = 2x^2 - 3x; x = 0; x = 3 for a, i know it goes from 0 to 1/2 but how should i write the equation? 0 to 1/2 INT x^3 + x or something?  | ||
|
ZpuX
Sweden1230 Posts
uhm or are you going to make integrals on them? | ||
Aesop
Hungary11304 Posts
On the first and third the interval is given. On the 2nd you have to find the place where they interject (but that is easy to calculate). The calculation process is very simple, find the upper curve, calculate the integral of the upper and substracte the integral of the lower in that section. If they vary, just divide the whole and calculate the different sections. And the functions you are dealing with are not really hard to integrate. | ||
|
Zherak
Norway256 Posts
For all three: Let a and b be the intersection between the two graphs given by y = (...); that is, solve y = x^2 + 4 = 6 - x for x. Find the area by integrating Y from a to b, where Y is one function of x subtracted from the other function of x. Depending on how which function you chose to subtract from which, and which intersectino you integrate from and two, you may very well end up with either a positive or a negative answer. The area is obviously the absolute value. For a): x^3 = x -> x = 0, x = 1 (let a = 0, b = 1) Y = x^3 - x +/-A = [int]a->b Y = [x^4/4 - x^2/2]0->1 = -1/4 A = 1/4 Here I have used the intersections as the limits of integration, which is what you should have done if they had not specified a = 0, b = 1/2. | ||
|
Raithed
China7078 Posts
| ||
|
Aesop
Hungary11304 Posts
On June 09 2008 01:21 Raithed wrote: so for a. its from x blah to x blah, but for the y given values, it is a line and a quadratic, right? I am not sure if I understand the question, though I would like to help. Maybe you mean this: y = x is a linear function ("line"). y = x^3 is of degree 3, so it rises more quickly ("quadratic", although cubic would be more accurate). Or what? But this has little to do with solving the task. The main obstacle here is properly integrating these functions. | ||
|
Raithed
China7078 Posts
a. y = x^3; y = x; x = 0; x = 1/2 x = blah and x = blah will be from "0" to "1/2" right? so i draw two dots essentially on the x-axis? thats what i mean, and for the y, i can plug in y = x^3 and y=x on the graphing calculator to solve, right? | ||
|
ZpuX
Sweden1230 Posts
 full of stupid mistakes | ||
|
ZpuX
Sweden1230 Posts
On June 09 2008 01:56 Raithed wrote: sorry if im unclear, given that a gives: a. y = x^3; y = x; x = 0; x = 1/2 x = blah and x = blah will be from "0" to "1/2" right? so i draw two dots essentially on the x-axis? thats what i mean, and for the y, i can plug in y = x^3 and y=x on the graphing calculator to solve, right? X = 0 and X = 1/2 is two vertical lines... Not sure what you want to do :S but yes, sketch the y = x^3 and y = x on the calculator and then draw the two vertical line, at x = 1/2 and x = 0 (will be the y-axis) the area enclosed by all the 4 equations (y = x^3 y = x x = 0 x= 1/2) will be the area you are looking for. | ||
|
Aesop
Hungary11304 Posts
On June 09 2008 00:19 Zherak wrote: Here I have used the intersections as the limits of integration, which is what you should have done if they had not specified a = 0, b = 1/2. On June 09 2008 01:59 ZpuX wrote: You are wrong tho... For a): x^3 = x -> x = 0, x = 1 (let a = 0, b = 1) Y = x^3 - x +/-A = [int]a->b Y = [x^4/4 - x^2/2]0->1 = -1/4 A = 1/4 this is not right. He wants to find the area that's enclosed by all the 4 lines, which makes it go from x = 0 --> x = 1/2. Zherak was just using the interjections as an example that would not spoil the exercise  | ||
|
Siefu
Australia205 Posts
On June 09 2008 02:03 ZpuX wrote: X = 0 and X = 1/2 is two vertical lines... Not sure what you want to do :S but yes, sketch the y = x^3 and y = x on the calculator and then draw the two vertical line, at x = 1/2 and x = 0 (will be the y-axis) the area enclosed by all the 4 equations (y = x^3 y = x x = 0 x= 1/2) will be the area you are looking for. Don't you just do the integral of x^3 minus the intergral of x from 0 to 1/2? | ||
pachi
Melbourne5338 Posts
| ||
|
Aesop
Hungary11304 Posts
On June 09 2008 02:17 Siefu wrote: Don't you just do the integral of x^3 minus the intergral of x from 0 to 1/2? Yes, those lines are just defining the interval in a less abstract way. | ||
|
blabber
United States4448 Posts
On June 09 2008 02:17 Siefu wrote: Don't you just do the integral of x^3 minus the intergral of x from 0 to 1/2? you mean integral of x - x^3 Raithed, imagine it like this. You have a square. Inside there's a smaller square. Now I ask you to find the area between the smaller square and the bigger square. What would you do? You find the area of the bigger square plus the smaller square, then subtract the area of the smaller square. It's basically the same thing here. Imagine y = x as the bigger square, then y = x^3 as the smaller square. Then you're doing this from x = 0 to x = 1/2. So it should be integral of x - x^3. (Remember, all integral is is just area under the curve) You can apply the same logic to the other problems. The MOST IMPORTANT THING TO SOLVING THOSE IS DRAWING THEM. By drawing them, you can find which equation is subtracted from which, and what the limits are | ||
|
Kwidowmaker
Canada978 Posts
I love you. | ||
|
ZpuX
Sweden1230 Posts
On June 09 2008 02:16 Aesop wrote: Zherak was just using the interjections as an example that would not spoil the exercise oh okay, sorry Zherak | ||
|
Raithed
China7078 Posts
On June 09 2008 02:03 ZpuX wrote: X = 0 and X = 1/2 is two vertical lines... Not sure what you want to do :S but yes, sketch the y = x^3 and y = x on the calculator and then draw the two vertical line, at x = 1/2 and x = 0 (will be the y-axis) the area enclosed by all the 4 equations (y = x^3 y = x x = 0 x= 1/2) will be the area you are looking for. this is what i wanted to know, thank you. no need for more complicated stuff guys! | ||
|
fight_or_flight
United States3988 Posts
On June 09 2008 02:20 pachi wrote: raithed, you can use http://www.texify.com/ for equations if you want them easier read You really should do this raithed. Just go to a wikipedia page with equations similar to the ones you need, click 'edit' and all the latex code for the equations are right there. Then you can just change them a little to what you need. | ||
| ||
|
|
BSL 21
Replay Cast
BASILISK vs Shopify Rebellion
Team Liquid vs Team Falcon
OSC
CrankTV Team League
Shopify Rebellion vs Team Liquid
BASILISK vs Team Falcon
Replay Cast
The PondCast
CrankTV Team League
Replay Cast
WardiTV Invitational
MaNa vs Gerald
Rogue vs GuMiho
ByuN vs Spirit
herO vs Solar
CrankTV Team League
[ Show More ] |
|
|
EPT
