Question

How do you integrate?

`int_0^4abs(x^2-4x+3)dx`

Answer 1

`int_0^4|x^2-4x+3|=4`

Explanation:

Given that we have an absolute valued integrand, the most important thing to determine is where in the interval of integration does the integrand change signs and split up our integral around the relevant points.

Factor `x^2-4x+3,` yielding `(x-3)(x-1)`, solving results in `x=3, x=1`.

From `[0, 1), x^4-4x+3>0,` so `|x^4-4x+3|=x^2-4x+3`.

So, one of our non-absolute valued integrals will be

`int_0^1(x^2-4x+3)dx=(1/3x^3-2x^2+3x)|_0^1=1/3-2+3=1/3-6/3+9/3=4/3`

Now, from `(1, 3), x^2-4x+3<0,` so `|x^4-4x+3|=-(x^4-4x+3)=-x^2+4x-3`

The second non-absolute valued integral will be

`int_1^3(-x^2+4x-3)=(-1/3x^3+2x^2-3x)|_1^3=-9+18-9+1/3-2+3=1/3-2+3=1/3-6/3+9/3=4/3`

Finally, on `(3, 4], x^2-4x+3>0, |x^2-4x+3|=x^2-4x+3,` meaning the final non-absolute value integral will be

`int_3^4(x^2-4x+3)=(1/3x^3-2x^2+3x)|_3^4=64/3-32+12-9+18-9=64/3-60/3=4/3`

So, adding up all the integrals, we get `4/3+4/3+4/3=12/3=4`