f(x) =2px, g(x) = 2x−4, and the y-axis.
For some reason I get -13.333. I did 4 above 0 (the points of intersection) integral { 2sqrt(x) - (2x-4) dx? Did I do anything wrong?
2006-08-15 09:10:59 · 4 answers · asked by ooeookillertofu 1 in Science & Mathematics ➔ Mathematics
Lets see, I'm assuming you meant that f(x) = sqrt(x), and the x-axis? Otherwise, none of it makes sense.
Look at the graph of the area. You need to do this in two parts. You should wind up with 20/3.
Edit: Actually, the y-axis is a suitable boundry, in which case your answer is 32/3.
2006-08-15 09:22:05 · answer #1 · answered by a_liberal_economist 3 · 0⤊ 0⤋
I assume f(x) = 2 sqrt x and g(x) = 2x - 4
Intersection:
2 sqrt x = 2x - 4
sqrt x = x - 2
x = x^2 - 4x + 4
(x - 1) (x - 4) = 0
x = 1 or x = 4 but x = 1 is extraneous.
INT(0,4) [g(x) - f(x)] dx
=
INT(0,4) [2x - 4 - 2 sqrt x]dx
=
x^2 - 4x - 4/3 x sqrt x | (0,4)
=
16 - 16 - 32/3 = 32/3, approx. 10.7
2006-08-15 16:28:16 · answer #2 · answered by dutch_prof 4 · 0⤊ 0⤋
A triangle?
What does 2px mean?
2006-08-15 16:36:36 · answer #3 · answered by Anonymous · 0⤊ 1⤋
yes...
2006-08-15 16:20:44 · answer #4 · answered by wizard 4 · 0⤊ 1⤋