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The functions f and g are given by f(x)=√x and g(x)=6x. Let R be the region bounded by the xaxis 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
 one year ago
 one year ago
The functions f and g are given by f(x)=√x and g(x)=6x. Let R be the region bounded by the xaxis 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
 one year ago
 one year ago

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Bladerunner1122Best ResponseYou've already chosen the best response.0
The functions f and g are given by f(x)=√x and g(x)=6x. Let R be the region bounded by the xaxis 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 yaxis 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.
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Can you upload a picture, because I can't open the link
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Well, I will upload a picture
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Painted area you need to find.
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Firstly, you need to find x values, where these two graphis intersects. Do you know how to do it?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
http://www.freeimagehosting.net/wzxvs
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
No, how do I do that?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
And if possible it should be without the calculator.
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
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=6x
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Well, it is easy, you won't need to use a caculator
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Secondly, you need to find there g(x)=6x intersects Ox line, do you know why you need to do it?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
Wouldn't g(x)=66?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
Since I have the point I don't need to solve for it right?
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
no, g(x) intererects Ox when 0=6x x=6
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
Mhmm... that make sense so far.
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
dw:1362062739064:dw So, R consist of two parts: R1 and R2, right?
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
If you want to find R=R1+R2, right?
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Area of R1: \[A=\int\limits_{0}^{4}\sqrt{x}dx\]
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Can you find R1 area for me?
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
No. Show me how you solve it?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
(2x^3/2)/3 + C
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
No, \[\int\limits_{0}^{4}\sqrt{x}dx=(\frac{ x ^{1,5} }{ 1,5 })4=\frac{ 4^{1,5}}{ 1,5}\]
 one year ago

KamilleBest ResponseYou've already chosen the best response.1
Oh, we got the same answer, ahh, sorry! So, now you have area of R1, do you know how to get area of R2?
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
So now I'm a little confused. Somehow your work doesn't look like my answer? And R2 would be the integral of 6x dx on the interval [4,6]?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
dw:1362064551772:dw \[\Large \int_0^2 (x_Rx_L) dy\] \[xR: y=6x\Rightarrow x=6y\\xL:y=\sqrt{x}\Rightarrow y^2=x\]
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
For the area of R2 I get 2. So the area for the total should be 16/3 + 2? = 22/3?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
dw:1362064821410:dw\[\Large V=\int_0^2A(y) \mathrm dy\]
 one year ago

Bladerunner1122Best ResponseYou've already chosen the best response.0
That reasoning doesn't make sense to me. I'm trying to calculate the area bounded by the x axis and the lines.
 one year ago
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