area of a region (integration)

[SilentSymphony]

whenever you are finding the area of a region enclosed by 2 graphs, can the resulting answer be negative? the only reason i ask is because ive been through this problem 3 times now looking for errors, and i cant find any, but every time, i still get -4.5 as the answer. here's my work:

"Find the area of the region enclosed by the graphs of y= x^2 -4x + 3 and y= x-1"

x^2 - 4x +3 = x-1

x^2 - 5x + 4 = 0

(x-4) (x-1) = 0

x= 4, x = 1

lim [4,0] of [(x^2 -4x+3)- (x-1)]dx

lim [4,0] of [x^2 -5x + 4] dx

1/3 x^3 - 5/2x^2 + 4x] [4,1]

64/3 - 40 + 16 - (1/3 - 5/2 + 4)= -4.5

any help at all would be great

 

[soroban]

Hello, SilentSymphony!

Did you make a sketch?


You work is correct except . . .

. . the line is the upper border, the parabola is the lower border.

You should have had: \(\displaystyle \L\:\int^{\;\;\;4}_1\left[(x\,-\,1)\,-\,(x^2\,-\,4x\,+\,3)\right]\,dx\)

 

[SilentSymphony]

oh okay that makes a lot of sense. thanks!