Inverse functions + finding area via volumes -- HELP NEEDED (1 Viewer)

[blackops23]

Hi guys, here's the question:

Q. Consider the function f(x) = 0.5(arcsinx)

Find the area of the region bounded the curve, the x-axis and the line x=1

----------------------------
My solution (which is wrong)

http://imageshack.us/photo/my-images/810/inverse.jpg/#

(Zoom the browser in with ctrl + scroll if its too small)

The answer is (pi/4 - 1/2) u^2

Can someone please explain why I am wrong, and how to obtain the correct answer without integration by parts?

Help is much appreciated

 

[Hermes1]

Hi guys, here's the question:

Q. Consider the function f(x) = 0.5(arcsinx)

Find the area of the region bounded the curve, the x-axis and the line x=1

----------------------------
My solution (which is wrong)

http://imageshack.us/photo/my-images/810/inverse.jpg/#

(Zoom the browser in with ctrl + scroll if its too small)

The answer is (pi/4 - 1/2) u^2

Can someone please explain why I am wrong, and how to obtain the correct answer without integration by parts?

Help is much appreciated

you could find the area between the curve and the y-axis. which would be easier to integrate. then take the area of the rectangle and minus the area you found with respect to the y-axis. this will give you the desired area.

tell me if u want a whole solution, i can write it up

 

[Hermes1]

ok so first u draw the graph. at x = 1 y = pi/4

then find x as a function of y
ie x = sin 2y

integrate that with respect to y between 0 and pi/4
you should get a 1/2

then get the area of the rectangle ie pi/4 multipled by 1
which is just pi/4

now Area of rectangle - area with respect to y-axis = area with respect to x-axis. as can be seen on graph.
therefore Area = pi/4 -1/2

 

[hscishard]

Make a rectangle, then find the area bounded by curve and y axis
You tried that?

arj1 confirmed it works
oh wait, it's hermes1 now?

 

[Hermes1]

Make a rectangle, then find the area bounded by curve and y axis
You tried that?

arj1 confirmed it works
oh wait, it's hermes1 now?

yes its me arj1. i changed my username.

 

[blackops23]

you could find the area between the curve and the y-axis. which would be easier to integrate. then take the area of the rectangle and minus the area you found with respect to the y-axis. this will give you the desired area.

tell me if u want a whole solution, i can write it up

ahhhhhhhhhh ok thank you very very much I get it know

 

[Drongoski]

 

[Hermes1]

but he didnt want to use integration by parts and this is in the mx1 forum.

 

[Drongoski]

Didn't read that. So he doesn't want it done using IBP. Maybe I should delete my solution.

 

[Hermes1]

Didn't read that. So he doesn't want it done using IBP. Maybe I should delete my solution.

no leave it up. bcuz sometimes in 3 unit ive seen them do it by parts. but they usually guide you through it in 3U.