Question

Find the value of \(\displaystyle\int_{1}^{4}\big(|x-1|+|x-2|+|x-3|\big)\,dx\).

Hint:

Split at x=1,2,3 and simplify the sum of absolute values on each piece before integrating.

Answer 1

Step 1: Identifying sign-change points inside [1,4]:
The three break points are \(x=1,2,3\), all within \([1,4]\), splitting it into \([1,2],[2,3],[3,4]\).

Step 2: Writing the integrand explicitly on each piece:
On \([1,2]\): \(|x-1|=x-1,\ |x-2|=2-x,\ |x-3|=3-x\), sum \(=4-x\). On \([2,3]\): \(|x-1|=x-1,\ |x-2|=x-2,\ |x-3|=3-x\), sum \(=x\). On \([3,4]\): all three are \(x-1,x-2,x-3\), sum \(=3x-6\).

Step 3: Integrating each piece:
\(\displaystyle\int_1^2(4-x)dx=\Big[4x-\tfrac{x^2}2\Big]_1^2=(8-2)-(4-\tfrac12)=6-3.5=2.5\). \(\displaystyle\int_2^3 x\,dx=\Big[\tfrac{x^2}2\Big]_2^3=4.5-2=2.5\). \(\displaystyle\int_3^4(3x-6)dx=\Big[\tfrac{3x^2}2-6x\Big]_3^4=(24-24)-(13.5-18)=0-(-4.5)=4.5\).

Final Answer:
Total \(=2.5+2.5+4.5=9.5=\dfrac{19}{2}\).\[ \boxed{\dfrac{19}{2}} \]