Sketch the region in the first quadrant that is bounded by the graphs of y=x^3, y=4x, and 2x + y-3=0 that lies below straight lines. Find the area of this region.
2007-03-26 01:09:54 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hi,
If you're truly looking for the area under straight lines, you can find their point of intersection at (.5,2) and then find the area under the 4x curve from x = 0 to .5 and then find the area under y = -2x + 3 from .5 to 1.5. Those give you .5 and 1 respectively for a total area of 1.5 under both of those lines measured down to the X axis. We now must remove the area under the x^3 equation. It intersects the -2x + 3 equation at (1,1), so we can find the area under the x^3 equation from 0 to 1 and then the area under the -2x + 3 equation from x = 1 to 1.5. Those give you .25 and .25 respectively for a total area of .5 under both of those lines measured down to the X, which is to be subtracted from our previous total area of 1.5. Since 1.5 - .5 = 1. 1 is the total area enclosed by these equations in the first quadrant. I used the integral command on the TI83 calculator to compute these.
I hope that helps.
2007-03-26 01:36:16 · answer #1 · answered by Pi R Squared 7 · 1⤊ 0⤋
65
2007-03-26 08:13:47 · answer #2 · answered by Professsor Daniel 2 · 0⤊ 0⤋
65
2007-03-26 08:12:37 · answer #3 · answered by Anonymous · 0⤊ 0⤋
0.9254 approx
First, you find the intersection of the two straight lines. Then you have to use the area between two functions formula twice, one between 0 to the first intersection, and the second time between the intersection point of the lines to the intersection point of the 3-2x with the x^3.
2007-03-26 08:26:56 · answer #4 · answered by dustlessus_101 2 · 0⤊ 0⤋