Question

Find the area of the shaded region

y=(x+2)^(1/2)
y=(1/x+1)
x= 0, 2

Answer 1

You first have to know which one is on top

Answer 2

http://www4b.wolframalpha.com/Calculate/MSP/MSP49719g2d9759c103f7b000018d94f2g6165hbb5?MSPStoreType=image/gif&s=12&w=367&h=192&cdf=Coordinates&cdf=Tooltips

Answer 3

ok

Answer 4

Looking at the graph, we know that (x+2)^1/2 is on top

Answer 5

So, Top - Bottom
\[\int\limits_{0}^{2}(x+2)^{1/2}-(1/(x+1))\]

Answer 6

dx

Answer 7

\[(X+1)^-1\]

Answer 8

that is what (1/x+1) can be rewritten as right?

Answer 9

Exactly

Answer 10

so now at this point I find anti derivative of the integral right?

Answer 11

yes

Answer 12

but when I do i get some weird decimal as an answer, and not the original answer

Answer 13

\[\int\limits_{0}^{2}\sqrt{x+2} -\int\limits_{0}^{2}(x+1)^{-1}\]
u=(x+2)
du = dx
\[\int\limits_{}^{}\sqrt{u} du = 2/3 u^{3/2}\]

Answer 14

\[2/3 (x+2)^{3/2}\]

Answer 15

\[u=x+1\]
du=dx
\[\int\limits_{}^{}(1/u )du \]

Answer 16

\[\ln (x+1)\]

Answer 17

2/3(4)^(3/2)- ln(2)

Answer 18

whoa we were never introduced into using ln(x+1) as a integral

Answer 19

Integral of 1/x is ln[x]

Answer 20

I just received a email from a buddy of mine and end up using this integral in the beginning

*next post*

Answer 21

\[\int\limits_{0}^{6} 2x - (x^2-4x)dx\]

Answer 22

now i am completely lost on where he gotten that from

Answer 23

me neither

Answer 24

I mean it would seem to correct to use this approach the way that we are doing the integral, which seems reasonable, but the fact that rewrote the integral into something different confuses me

Answer 25

I have no idea where this integrand came from

Answer 26

Can the integral be rewritten in order to get a whole number as a answer, or it impossible to do so with only knowing Calculus 1?

Answer 27

you get same number no matter how you set up the integral. Sometime they are set up one way or another to make it easier to integrate

Answer 28

Forgot that I can use U substitution in order to use these integrals so I can figure this out a lot faster.

So, I just use this approach and ask my buddy where did he even think about using a different integral.

Answer 29

Thanks for the help though. I appreciated it