f(x) is always above the x axis in that interval. Therefore just integrate the function from 1 to 4. I don't know what you mean by "the limit process".
The indefinite integral is x^4/2 -x^3/3 +x. That equals 128 -21.33 +4 = 110.67 at x = 4 and 0.50 -0.33 +1 = 1.17 at x = 1.
The area should be 110.67-1.17 = 109.50
My algebra tends to be sloppy, so my answer could be wrong, too.
f(x)=2x^3-x^2+1 [1,4]=interval
Set up the limit process (once with right endpts, and the other with left entdpts.) to find the area under the curve and above the x-axis on the given interval
Should the right and left endpt limit set ups have the same area? If so what is it?
PLEASE HELP.....I GOT 107.55 BUT IT IS NOT THE RIGHT ANSWER AND I DON'T SEE HOW THERE CAN BE A DIFFERENCE IN SETTING UP
2 answers
thanks so much!!!