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Can you upload a picture, because I can't open the link

|dw:1362062051357:dw|

Painted area you need to find.

Firstly, you need to find x values, where these two graphis intersects. Do you know how to do it?

http://www.freeimagehosting.net/wzxvs

No, how do I do that?

And if possible it should be without the calculator.

Well, it is easy, you won't need to use a caculator

Secondly, you need to find there g(x)=6-x intersects Ox line, do you know why you need to do it?

Wouldn't g(x)=6-6?

Since I have the point I don't need to solve for it right?

no, g(x) intererects Ox when 0=6-x x=6

|dw:1362062632093:dw|

Mhmm... that make sense so far.

|dw:1362062739064:dw|
So, R consist of two parts: R1 and R2, right?

If you want to find R=R1+R2, right?

Area of R1:
\[A=\int\limits_{0}^{4}\sqrt{x}dx\]

Can you find R1 area for me?

16/3?

No. Show me how you solve it?

(2x^3/2)/3 + C

No,
\[\int\limits_{0}^{4}\sqrt{x}dx=(\frac{ x ^{1,5} }{ 1,5 })|4=\frac{ 4^{1,5}}{ 1,5}\]

For the area of R2 I get 2. So the area for the total should be 16/3 + 2? = 22/3?

|dw:1362064821410:dw|\[\Large V=\int_0^2A(y) \mathrm dy\]