Question

Find the area bounded by the x-axis, y= sqrt(x-2), and y=-x

Answer 1

first we need to find the limits if integration, we do this be setting the equations equal to each other

Answer 2

I found that this needs to be integrated by parts (if you integrate horizontally). The first limits are from -2 to 0, and the other limit is from 0 to 2

Answer 3

okay so simply integrate \[(\sqrt{x-2})-(-x)\], using those limits

Answer 4

For the first area between -2 and 0, I got \[4^{3/2} / 3\] I was wondering if you get the same. This was a question that we did in class, and I got an answer different from the prof. Also, I used U-substitution and ended up changing the limits of integration. The prof. did the problem a different way that I've never seen before. So really, I'm just wondering if you get the same answer for that first area.

Answer 5

i am not in a position to do the computation right now, but let me get you someone who can alright

Answer 6

k

Answer 7

hold the phone

Answer 8

there is no solution to this. you have
\[y=\sqrt{x-2}\] and
\[y=-x\] if i read it correctly. these do not intersect

Answer 9

They intersect at (-2,2). But, this is also bounded by the x-axis. Because of this, you need to integrate by parts, or integrate with respect to y (which I'm not supposed to do yet)

Answer 10

no they don't

Answer 11

oh crud, I meant sqrt(2-x). I'm sorry I didn't catch that

Answer 12

in particular you cannot evaluate
\[f(x)=\sqrt{x-2}\] at -2 you will get
\[\sqrt{-4}\]

Answer 13

oooooooooooooooooh ok maybe we can do it now

Answer 14

now they intersect at -2

Answer 15

so you have to break up the integral into two pieces right?

Answer 16

because you are above the x - axis. so first take
\[\int_{-2}^0(\sqrt{2-x}+x)dx\]then
\[\int_0^2\sqrt{2-x}dx\]

Answer 17

first anti derivative is
\[-\frac{2}{3}(2-x)^{\frac{3}{2}}+\frac{x^2}{2}\] second is just
\[-\frac{2}{3}(2-x)^{\frac{3}{2}}\]

Answer 18

now plug in limits and be done

Answer 19

My teacher said we needed to use u-substitution

Answer 20

fine. you can use
\[u=2-x\]
\[du=-dx\]
\[-du=dx\] and get
\[\int\sqrt{2-x}dx=-\int\sqrt{u}du=-\frac{2}{3}u^{\frac{3}{2}}\]

Answer 21

that is what i did but i didn't make it explicit. only off by a minus sign that is why

Answer 22

now replace u by 2-x and get the answer i wrote. (i hope)

Answer 23

alright, that's what I was looking for. Thanks!

Answer 24

yw