Integration question pls help

Evaluate ∫|x^2-4x+3|dx from 0 to 2 without the use of a graphic calculator. Pls show clear workings. Answer is 2.

Integrating an absolute value confuses most simply because we have no intuition as to what it really means. An absolute value is simply a continuous piecewise function, and we can show that with |x^2-4x+3|

Let f(x) = |x^2-4x+3| Find the zeros: (x - 3)(x - 1) = 0 , x = 3 and 1

We can say that f(x) = -(x^2 - 4x + 3) when 1 (x^2 - 4x + 3) For all other x.

So now that we have split this into a continuous piecewise function, we can now see which of the two, and it can be both as it is in this case, applies to our limits of integration. We are integrating from 0 to 2, so we also split the integral into:

∫x^2-4x+3 dx from 0 to 1 and into ∫-(x^2-4x+3) dx from 1 to 2

= (x^3/3) - (2x^2) + 3x evaluated from 0 to 1 and -(x^3/3) + (2x^2) - 3x evaluated from 1 to 2.
= 1/3 - 2 + 3 = -5/3 + 3 = 4/3
(-8/3 + 8 - 6) + 4/3 = -2/3 + 4/3 = 2/3
Add them together and we get: 4/3 + 2/3 = 2, and thus our answer.

Complete the square to help you analyze the function better:

x^2 - 4x + 3 = (x - 2)² - 1

So from analyzing the above expression, we can see that the function will be negative between (1,2] and positive between [0,1]. So therefore, since absolute value shifts area underneath the x-axis to become positive area above the x-axis, the area is:

∫(x - 2)² - 1 dx from 0 to 1 + ∫(1 - (x - 2)²) dx from 1 to 2

= (x - 2)^(3)/3 - x eval. from 0 to 1 + x - (x - 2)^(3)/3 eval. from 1 to 2

= 4/3 + 2 - 1 - 1/3 = 6/3 = 2

|x^2-4x+3| = |(x-1)(x-3)|
for x between 0 and 1 expression |(x-1)(x-3)| becomes x^2 - 4x + 3
for x between 1 and 2 expression |(x-1)(x-3)| becomes - x^2 + 4x - 3
∫|x^2-4x+3|dx from 0 to 2 =
∫(x^2-4x+3)dx from 0 to 1 + ∫(- x^2+4x-3)dx from 1 to 2 =
[ (1/3)x^3 - 2x^2 +3x] from 0 to 1 + [ -(1/3)x^3 +2x^2 -3x] from 1 to 2 =
[ (1/3)1^3 - 2*1^2 +3*1] - [ (1/3)0^3 - 2*0^2 +3*0] +
[ -(1/3)2^3 +2*2^2 -3*2] - [ -(1/3)1^3 +2*1^2 -3*1] =
(1/3) +1 - (8/3) +2 + (1/3) +1 = 2

x² – 4x + 3 = (x – 3)(x – 1), so the quadratic is negative from x = 1 to x = 3. Integrate the absolute value as follows

F(x) = ∫x² – 4x + 3 dx = x³/3 – 2x² + 3x

∫x² – 4x + 3 dx [0, 2] = ∫x² – 4x + 3 dx [0, 1] – ∫x² – 4x + 3 dx [1, 2] =

(F(1) – F(0)) – (F(2) – F(1)) = 2F(1) – F(2) – F(0) = 8/3 – 2/3 – 0 = 6/3 = 2