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April 25, 2014

April 25, 2014

Posted by **Mishaka** on Wednesday, February 29, 2012 at 2:39pm.

(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**, Wednesday, February 29, 2012 at 3:10pmSince y^2 = x is not even defined left of the y-axis, 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**, Wednesday, February 29, 2012 at 3:11pmfind 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**, Wednesday, February 29, 2012 at 3:12pm33/2 is correct.

- Calculus (Area Between Curves) -
**Mishaka**, Wednesday, February 29, 2012 at 3:35pmThank 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 y-axis. So, I, too, figured that 33/2 would be the most reasonable answer. Thank you both!

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