Show your integration.
I would hate to find out that your answer was (log 8)/2
Find the area in the first quadrant which is bounded by xy=1, xy=4, y=x and y=2x
I attempted the question by letting u=xy and v= y then i got the equations x=u/v and y=v
I then found the jacobian which was equal to 1/v
0<u<4 and x<v<2x
I thought the limits for x was 0<u<4 and x<v<2x but the problem is the limits for the x and 2x i wasn't sure how to integrate it when the double integral was in terms of dvdu i tried using the formula x=u/v to change the limits for x and 2x but it didnt give me the right answer...which was suppose to be (3log2)/2
No the answer i got was 4log2 which is why i think i must have made a mistake somewhere
I had the integral from 4-0 integral 2u/v to u/v of 1/v dvdu. when integrated i got {logv} from 2u/v and u/v which gave me log(2u/v) - log(u/v) which equals log((2u/v)/(u/v)) = log2. Then i have integral from 4-0 log2 du which should just be {ulog2}from 4-0 which is 4log 2 but I think i'm doing it wrong...
Easiest way to approach this problem is to break it up into two parts.
First draw the graphs together with a calculator of my favorite, MATLAB.
You can see that the area you want is bounded above by xy=4 and below by xy=1, but it is also bounded by y=x and y=2x.
If you drop a perpendicular from the point where xy=4 intersects y=2x, you can esentially divide this problem into two solvable areas.
You first figure the area under y=2x then subtract the area under xy=1.
Now you find the x boundaries by solving xy=1 so y=1/x then substitute into y=2x and you get 1/x = 2x or .
Do the same for y=2x and xy=4 so you get or .
Now integrate y=2x from the lower bound to the upper bound and do the same for y=1/x. The subtract the latter from the former.
Now you do the same for the second half of the area.