Jawabanmu

2013-12-13T22:44:22+07:00
x^2-4x+3=0
(x-1)(x-3)=0
x=1x=3

jadi 
L= -\int\limits^3_1 {(x^2-4x+3)} \, dx =\int\limits^1_3 {(x^2-4x+3)} \, dx

ada minesnya krn klo digambar, letaknya ada dibawah sumbu x
= \frac{1}{3} x^3-2x^2+3x ]_3^1

= (\frac{1}{3} 1^3-2.1^2+3.1) -(\frac{1}{3} 3^3-2.3^2+3.3)

= (\frac{1}{3}-2+3) -(9-18+9)

= \frac{1}{3}-2+3 -9+18-9

= 1\frac{1}{3}= \frac{4}{3}