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

Question

Emily Ammerman · Accepted Answer

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

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#

Scarlett Allison · Answer

undefined To evaluate the definite integral $ \int_{0}^{4} |x^2 - 4x + 3| \, dx $, we first need to find the critical points of the absolute function $ |x^2 - 4x + 3| $. The critical points occur where the function inside the absolute value changes sign, which happens at $ x = 1 $ and $ x = 3 $.

Split the integral into two parts based on these critical points:

1. From $ x = 0 $ to $ x = 1 $:
   $ \int_{0}^{1} (x^2 - 4x + 3) \, dx $

2. From $ x = 1 $ to $ x = 3 $:
   $ \int_{1}^{3} -(x^2 - 4x + 3) \, dx $

3. From $ x = 3 $ to $ x = 4 $:
   $ \int_{3}^{4} (x^2 - 4x + 3) \, dx $

Evaluate each of these integrals separately using the power rule for integration:

1. $ \int_{0}^{1} (x^2 - 4x + 3) \, dx = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{0}^{1} = \frac{1}{3} - 2 + 3 - 0 = \frac{4}{3} $

2. $ \int_{1}^{3} -(x^2 - 4x + 3) \, dx = -\left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{1}^{3} = -\left( \frac{27}{3} - 18 + 9 - \frac{1}{3} + 2 + 3 \right) = -\frac{32}{3} $

3. $ \int_{3}^{4} (x^2 - 4x + 3) \, dx = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{3}^{4} = \left( \frac{64}{3} - 32 + 12 - \frac{27}{3} + 18 - 9 \right) = \frac{64}{3} $

Summing these values gives the final result:

$ \int_{0}^{4} |x^2 - 4x + 3| \, dx = \frac{4}{3} - \frac{32}{3} + \frac{64}{3} = \frac{36}{3} = 12 $