Take the integral of these two functions: x^3 and 4x from 0 to the point where 3 - 2x and 4x intersect (1/2). Subtract your answers.
Do the same from 1/2 to the point where 3 - 2x = x^3 (1)
The region R in the first quadrant that is bounded by the graphs of y=x^3, y=4x, and y=-2x+3=0 and that lies below both straight lines. Find the area of this region.
I tried to find the area of the bigger triangle-the smaller region:
I took the integral of -6x+3 from 0 to 1.5 and got -2.25
then i took the integral of -x^3-2x+3 and got .984375
Altogether I get -3.234375. Is this right?