Kenneth S. answered • 12/15/16
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The domain of y = √(2x) is restricted to the first quadrant x values.
This concave-downward function intersects the line at x=2 and the enclosed area consists of
A1, above x-axis and beneath the square root curve, on [0,1] and then
A2 which is beneath the square root curve but above the line, on [0,2].
Check all of these facts before proceeding to do the integrals.
A1 = ∫01 2½x½dx = (2/3)√2 ... def. integral from 0 to 1...
A2 = def. integral from 1 to 2 of (2½x½ -2x + 2) dx ... and when I computed A2 + A1 I got 5/3.
It's pretty hideous to try to type all of the intermediate steps, so I'll just post this and then check my work.
You should do the work, too, and IF I FIND THAT I'VE A MISTAKE TO REPORT, I WILL POST A COMMENT.
Kenneth S.
Earlier I posted the answer as 17/3 which is 5 and one third.
That could not be right because is you look at the square in the first quadrant with vertices (2,0), (0,2) (2,2) and(0,0) then that would be 4 square units, which is more than the enclosed area. After careful work ON PAPER, rather than on screen, I've come up with answer 5/3, which seems plausible.
If you have a graphing calculator that can do definite integrals, that's a very good way to check definite integrations (remember to do two separate ones & sum them).
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12/15/16
Kenneth S.
Bad answer was 17/3; bad comment by me --- that bad answer is 5 2/3. Having an error prone day!
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12/15/16
Samson P.
12/15/16