**The Area and Volume of bounded functions:**

This is not the most informative blog, but for those who have a rough idea of whats going on should be able to grasp and understand the examples shown.

Find the area between

y = 2 - x^2 and y = (1/2)x.

from x = 0 to x = 1 .

Now as seen above we use integration to determine this. Basically you find the integral of the equation shown and solve for the domain of it. Which in this case is x = 0 to x = 1.

Now the example jumps right to the answer being 17/12. The actual steps behind the seen is

= [ 2x - (x^3 / 3) - (x^2 / 4)] from x = 0 to x = 1

=[ (2(1) - (1^3 / 3) - ( 1^2 / 4) - (2(0) - (0^3)/3 - 0^2 / 4) ]

= [ 2 - 1/3 - 1/4 - 0 - 0 - 0]

= [ 2 - 1/3 - 1/4]

= [ 6/3 - 1/3 - 1/4] = [ 5/3 - 1/4] = [20/12 - 3/12] = 17/12 units squared. YAH!

So after these steps we have found the area between these two functions from x = 0 to x = 1.

**Find the area:**

y = x^(1/3) , and y = x^3 from x = 0 to x = 2.

First thing we should notice about this is that they intersect at x= 1. Therefore we need to split this up into two different integrals. One from x = 0 to 1, and the other from x = 1 to 2.

= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 0 to x = 1

= [ 3/4 (1) ^(4/3) - 1/4 (1) ^4 - 3/4 (0) ^(4/3) - 1/4 (0) ^4

= 3/4 - 1/4 - 0 - 0 = 3/4 - 1/4 = 2/4 = 1/2 YAH!

= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 1 to x =2

= [ 3/4 (2) ^(4/3) - 1/4 (2) ^4 - 3/4 (2) ^(4/3) - 1/4 (2)^4]

= 4 - 3/2(2)^1/3 - 1/4 - 3/4

= 4.5 - 1.5(2)^1/3

So add up the values of both these integrals to get

. 5 + 4.5 - 1.5(2)^1/3 = 3.11 approximate units squared.

This is how you find the area between two functions bounded.

For the first time: I know what you're talking about :D

ReplyDeletei have no idea what i just read :/

ReplyDeletei don´t get it^^

ReplyDeleteinteresting

ReplyDeleteoh lawd lol

ReplyDeletewtf man

ReplyDeleteEasy stuff, but if only the bastards up in bloggerland can make it so the ^2 thing makes a little 2 above the number. Shit gets confusing after a bit.

ReplyDeleteYour explanation is clear however I know how to do this already. Could you tell something about derivativing sin(x) and cos(x) etc.? following

ReplyDeleteWhat did I just read? Its like, wow.

ReplyDeleteThat was confusing, thx though!

ReplyDeleteMy brain is about to explode.

ReplyDeleteThanks! Never really have figured this out before :)

ReplyDeleteBrain just got punted through a field goal.

ReplyDeleteMy brain hurts now, but thanks for giving us the solution to that problem!

ReplyDeleteOh man, the first I understand something here! This brings back some good memories. :p

ReplyDeleteconfusing?? yes

ReplyDeleteI think it'd be easier to learn chinese...

ReplyDeleteI remember doing this. Thinking back on it I'm glad I was doing that instead of some of the other stuff I've seen.

ReplyDeleteand this is why i'm not majoring in math :)

ReplyDeleteHey man next time I´ll send you my homework, ok with you? Just kiddin´dude haha.

ReplyDelete