y=f(x)=x²-4x+3=(x-1)(x-3).
So the x-intercepts are at x=1 and x=3, giving us the limits of integration. The curve sits under the x-axis between these two points because, if we choose a value of x between 1 and 3, say 2, then f(2)=4-8+3=-1, which is below the x-axis.
This gives us a mental picture of the curve.
So the integral is:
∫₁³f(x)dx=
∫₁³(x²-4x+3)dx=
[x³/3-2x²+3x]₁³=
(9-18+9)-(⅓-2+3)=0-1⅓=-4/3.
It’s negative because the area is below the x-axis, so the area is 4/3.