Question

find the area enclosed by the x axis and the curve

x= t^3+1 y= 2t-t^2

Answer 1

@zepdrix

Answer 2

how do i find my bounds ? i totally forgot lol

Answer 3

For parametric equations?
Hmm that's a good question :d

Answer 4

yeah :O

Answer 5

One of the boundaries is the x-axis...
The x-axis is the line y=0.

So our curve will intersect the x-axis when y=0.\[\large\rm y=2t-t^2\]\[\large\rm 0=2t-t^2\]Solve for t.

Answer 6

i got t= 0 and t= -2

Answer 7

-2? Hmm that doesn't seem right :d

Answer 8

Check again lady.
t(2-t)=0

Answer 9

oh wait lmao t= 0 and t= 2

Answer 10

Okieeee, so those are probably the boundaries we'll use.
I'm trying to remember how to find area when given a parameter like this...
sec hold upppp :)

Answer 11

okay to find our bounds we solve for t?

Answer 12

Yes. That's what we did.

Answer 13

okie lol

Answer 14

So I guess our formula for area is,\[\large\rm A=\int\limits_{t_1}^{t_2}y~dx\]We found our boundary values,\[\large\rm A=\int\limits\limits_{0}^{2}y~dx\]And we know what y is,\[\large\rm A=\int\limits\limits\limits_{0}^{2}(2t-t^2)~dx\]Let's try to find the differential of x.

Answer 15

\[\large\rm x=t^3+1\]\[\large\rm dx=?\]

Answer 16

3t^2

Answer 17

Yes,\[\large\rm dx=3t^2~dt\]So therefore our integral is,\[\large\rm A=\int\limits\limits\limits\limits_{0}^{2}(2t-t^2)(3t^2)dt\]

Answer 18

*I think*

Answer 19

oh okay so then we take one derivative right ?

Answer 20

We already did.
\(\large\rm x=t^3+1\qquad\to\qquad dx=3t^2~dt\)

Answer 21

oh yeahh lool

Answer 22

i got 24/5

Answer 23

Mmmm ya that's what Wolfram says also :O

Answer 24

So hopefully we did that properly :d

Answer 25

ʕっ•ᴥ•ʔっ
Bear says, "good job!"

Answer 26

loool

Answer 27

ʕ´• ᴥ •`ʔ