when y is less than one, y^3 is less than y^2, so you should have
INT (y^2-y^3) dy
Can someone look over my work and tell me if my steps look correct? I'm trying to correct some problems that looked wrong.
Instructions: Find the total areas between the given curves.
1. x= (y^3) and x=(y^2) on the interval [0,1]
(integral from 0 to 1 of) ((y^3)-(y^2))dy = (integral from 0 to 1)(y^3)dy - (integral from 0 to 1)(y^2)dy=
(y^4)/4 - (y^3)/3 from 0 to 1
(1^4)/4 - (1^3)/3= -1/12
but area can't be negative... did I subtract them in the wrong order?
2. y= (x^2) and x+y=2 on the interval [0,2]
y1= x^2
y2= 2-x
x^2=2-x
x^2+x-2=0
(x+2)(x-1)=0
x=-2, 1
A= (integral from 0 to 2)(2-x)dx - (integral from 0 to 1)(x^2)dx= 5/3
Thank you and I apologize in advance if my notation is confusing.
2 answers
Thank you!