Question

How do you evaluate the definite integral `int abs(x^2-4x+3)dx` from [0,4]?

Answer 1

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

Explanation:

graph{|x^2-4x+3| [-2, 5, -1, 5]}

We need to perform the integration in several steps because of the modulus sign

We want to evaluate `I=int_0^4 |x^2-4x+3| dx`

Note By symmetry about x=2 we just need to evaluate
`I=2int_0^2 |x^2-4x+3| dx`, and observe that

`|x^2-4x+3| = { (x^2-4x+3,0<=x<=1), (-(x^2-4x+3),1<=x<=2) :}`

Hence,
`1/2I=int_0^2 |x^2-4x+3| dx`
`:. 1/2I=int_0^1 |x^2-4x+3| dx + int_1^2 |x^2-4x+3| dx`
`:. 1/2I=int_0^1 (x^2-4x+3) dx + int_1^2 -(x^2-4x+3) dx`

`:. 1/2I= [1/3x^3-2x^2+3x]_0^1 -[1/3x^3-2x^2+3x]_1^2`

`:. 1/2I= (1/3-2+3) -{(8/3-8+6)-(1/3-2+3)}`
`:. 1/2I= 4/3 -{2/3-4/3}`
`:. 1/2I= 4/3 -(-2/3)`
`:. 1/2I= 4/3 +2/3`
`:. 1/2I= 2`
`:. I= 4`