(a) ∫022x−x2dx
=[x2−3x3]02
=4−38
=34
(b)
calculate half of the area found in part A
=34(21)=32
For the function y = mx, one of the points is the origin (0,0). we need to find the other point that intersects the curve. proceed as follows
mx=2x−x2x2+mx−2x=0x2+x(m−2)=0x=0 or x=2−m
find x value in terms of m where curves intersect
∫02−m(2x−x2−mx)dx=32
[x2−3x3−2mx2]02−m
(2−m)2−3(2−m)3−2m(2−m)3=32
(2−m)2[33−32+62m−63m=32
61(2−m)2[2−m]=32
61(2−m)3=32
Evaluate and solve m
(2−m)3=32(6)
(2−m)=34
m=2−34
Comments
Leave a comment