Question

checking my answer :) Find the area enclosed by the curves: y=x^2 -2x y=2x-x^2

Answer 1

My answer is 4

Answer 2

How did you get that answer?

Answer 3

found zeros: 0 and 2

Answer 4

points of interesction: 0,0 and 2,0

Answer 5

top function: 2x-x^2

Answer 6

bottom function: x^2 -2x

Answer 7

Yes, then the integral is not correct.

Answer 8

Which is your indefinite integral, before substitution?

Answer 9

well, yt-yb delta x=

Answer 10

2x-x^2-(x^2 -2x)

Answer 11

=2x-x^2 -x^2 +2x

Answer 12

\[\Large \int_{0}^{2} (4x-4x^2)~dx\]

Answer 13

yes, just caught my mistake

Answer 14

but I have -2x^2

Answer 15

Whoops, right, sorry, I just woke up >.<
But yea it's -2x^2, not 4.

Answer 16

cool well I separated 2 out so I have 2x-x^2 to evaluate

Answer 17

so its then 2[x^2 -(x^3)/3]

Answer 18

As in:
\[\Large 2\int_{0}^{2} (2x-x^2) dx\]?

Answer 19

yes

Answer 20

Right, now just evaluate at your limits
\[\Large 2(x^2-\frac{x^3}{3}|_{0}^{2})\]

Answer 21

ok so

Answer 22

8/3

Answer 23

It would appear so. Anyone have any objections?

Answer 24

:) alright, thanks for the check

Answer 25

You're welcome :P