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 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
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

This Question is Closed

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

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

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

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

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

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0No, how do I do that?

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0And if possible it should be without the calculator.

Kamille
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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

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

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

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't g(x)=66?

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0Since I have the point I don't need to solve for it right?

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

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0Mhmm... that make sense so far.

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

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

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

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

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

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0(2x^3/2)/3 + C

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

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

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0So 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]?

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0dw:1362064551772:dw \[\Large \int_0^2 (x_Rx_L) dy\] \[xR: y=6x\Rightarrow x=6y\\xL:y=\sqrt{x}\Rightarrow y^2=x\]

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0For the area of R2 I get 2. So the area for the total should be 16/3 + 2? = 22/3?

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0dw:1362064821410:dw\[\Large V=\int_0^2A(y) \mathrm dy\]

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0That 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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