Q.Evaluate the definite integral $\int_{1}^{4}\left[|x-1|+|x-2|+|x-3|\right]dx$

The key idea is that the absolute value function is piecewise linear, so we split the interval $[1,4]$ at the points where the expressions inside the absolute values change sign: $x=1$, $x=2$, and $x=3$.

Step 1: Write the piecewise definitions.

On $[1,4]$:

• $|x-1| = x-1$ (since $x \ge 1$).
• $|x-2| = \begin{cases} 2-x, & 1 \le x < 2 \\ x-2, & 2 \le x \le 4 \end{cases}$.
• $|x-3| = \begin{cases} 3-x, & 1 \le x < 3 \\ x-3, & 3 \le x \le 4 \end{cases}$.

Step 2: Split the integral into three subintervals.

Step 3: Evaluate each part.

• On $[1,2]$: integrand = $(x-1) + (2-x) + (3-x) = 4 - x$. $\int_{1}^{2} (4-x)\,dx = \left[4x - \frac{x^2}{2}\right]_{1}^{2} = (8-2) - (4-0.5) = 6 - 3.5 = 2.5$.
• On $[2,3]$: integrand = $(x-1) + (x-2) + (3-x) = x$. …

We need to evaluate

The key idea: an absolute value $|x-a|$ is a piecewise linear function — it equals $x-a$ when $x \ge a$, and $a-x$ when $x \le a$. So the whole integrand changes its algebraic form at each of the points $x=1$, $x=2$, and $x=3$. Since our integration limits are from $1$ to $4$, we must split the interval $[1,4]$ into subintervals where each absolute value has a fixed sign.

Let’s list the breakpoints in order: $1$, $2$, $3$. That gives us three subintervals to consider: $[1,2]$, $[2,3]$, and $[3,4]$. On each, we rewrite the integrand without absolute values.

1. On $[1,2]$:

   • $|x-1| = x-1$ (since $x \ge 1$)
   • $|x-2| = 2-x$ (since $x \le 2$)
   • $|x-3| = 3-x$ (since $x \le 3$; actually $x \le 2$ so definitely $x \le 3$)

   So the integrand becomes:

2. On $[2,3]$:

   • $|x-1| = x-1$
   • $|x-2| = x-2$ (since $x \ge 2$)
   • $|x-3| = 3-x$ (since $x \le 3$)

   Sum:

3. On $[3,4]$:

   • $|x-1| = x-1$
   • $|x-2| = x-2$
   • $|x-3| = x-3$ (since $x \ge 3$)

   Sum:

Now the original integral becomes the sum of three definite integrals:

Compute each:

• First integral: …

Method: Split the interval at every point where an absolute value changes sign

Use this for $\int_p^q \sum_i|x-a_i|\,dx$: each $|x-a_i|$ is piecewise linear, so break the domain at each $a_i$ and remove the bars on each subinterval.

Steps

Step 1: List the breakpoints inside the interval.

Every $|x-a|$ switches from $a-x$ (when $x\le a$) to $x-a$ (when $x\ge a$) at $x=a$. Collect all such $a_i$ that lie in $[p,q]$ and order them.

Step 2: Rewrite the integrand without bars on each subinterval. …

Common Mistakes

Mistake 1: Not splitting at every breakpoint.

Why it's wrong: each $|x-a|$ changes sign at its own $a$ (here $1,2,3$); splitting only once misrepresents the integrand. Correct approach: break $[1,4]$ at $2$ and $3$ (and note $1$ is an endpoint).

Mistake 2: Wrong sign on a subinterval. …