Posted by Mishaka on .
Find the area of the region bounded by the curves y^2=x, y4=x, y=2 and y=1
(Hint: You'll definitely have to sketch this one on paper first.) You get:
a.) 27/2
b.) 22/3
c.) 33/2
d.) 34/3
e.) 14

Calculus (Area Between Curves) 
Steve,
Since y^2 = x is not even defined left of the yaxis, the region is
x<3: a triangle bounded by y=x+4, x=3, y=2
area = 3*3/2 = 9/2
3<x<0: a rectangle, area 3*3=9
0<=x<=1: a curved triangular area
area = Int(1√x)dx [0,1]
= (x  2/3 x√x)[0,1]
= (1  2/3) = 1/3
Total: 9/2 + 9 + 1/3 = 14  1/6
Almost (e), but not quite. Typo somewhere? 
Calculus (Area Between Curves) 
Nade,
find the definite integral y^2  y + 4 evaluated from 2 to 1. I go 16.5 sq units Choice c

Calculus (Area Between Curves) 
Steve,
33/2 is correct.

Calculus (Area Between Curves) 
Mishaka,
Thank you Nade for making this correction! I went back and graphed the functions with pencil to see if I could figure a rough estimate for the area. I found that these function make an odd trapezoid shape. The area of the solid trapezoid portion of the graph is 15 and then there is a small portion on the other side of the yaxis. So, I, too, figured that 33/2 would be the most reasonable answer. Thank you both!