The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. h t t p://goo.gl/jXIZD

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The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. h t t p://goo.gl/jXIZD

Calculus1
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The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. h t t p://goo.gl/jXIZD 1. Find the area of R. 2. The region R is the base of a solid. For each y, where 0<=y<=2, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 2y. Write, but do not evaluate, an integral expression that gives the volume of the solid. 3. There is a point P on the graph of f at which the line tangent to the graph of f is perpendicular to the graph of g. Find the coordinates of point P.
Can you upload a picture, because I can't open the link
|dw:1362062051357:dw|

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Other answers:

Well, I will upload a picture
1 Attachment
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, actually, you don't need to, because you have a pont (4;2) given. If you don't have any point, you need to solve this: √x=6-x
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, \[\int\limits_{0}^{4}\sqrt{x}dx=(\frac{ x ^{1,5} }{ 1,5 })|4=\frac{ 4^{1,5}}{ 1,5}\]
Oh, we got the same answer, ahh, sorry! So, now you have area of R1, do you know how to get area of R2?
So now I'm a little confused. Somehow your work doesn't look like my answer? And R2 would be the integral of 6-x dx on the interval [4,6]?
|dw:1362064551772:dw| \[\Large \int_0^2 (x_R-x_L) dy\] \[xR: y=6-x\Rightarrow x=6-y\\xL:y=\sqrt{x}\Rightarrow y^2=x\]
For the area of R2 I get 2. So the area for the total should be 16/3 + 2? = 22/3?
|dw:1362064821410:dw|\[\Large V=\int_0^2A(y) \mathrm dy\]
That reasoning doesn't make sense to me. I'm trying to calculate the area bounded by the x axis and the lines.

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