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Bladerunner1122
 2 years 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
Bladerunner1122
 2 years 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

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Bladerunner1122
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Can you upload a picture, because I can't open the link

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

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

Kamille
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0No, how do I do that?

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

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

Kamille
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Wouldn't g(x)=66?

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

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

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

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

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

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

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

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

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

Bladerunner1122
 2 years 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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