Find the area enclosed between f(x)=x^2-4x+3 and the x-axis.

Question

I Don’t Know

Answer 1

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.