By using the properties of definite integrals, evaluate the integral $\int_{0}^{4}|x-1|\,dx$

Question

Accepted Answer

> [!TLDR]
> The integral $\int_{0}^{4}|x-1|\,dx$ is split at the point where the absolute value changes sign ($x=1$), turning it into the sum of two simple polynomial integrals. The final value is $5$.

The absolute value function is the classic case for using the **split property** of definite integrals. The core idea is simple: an absolute value creates a piecewise function — it behaves one way on one interval and another way on the next. You cannot integrate $|x-1|$ directly as a single expression because its rule changes at $x=1$.

The property we use is:
$$
\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx
$$
where $c$ is any point between $a$ and $b$. We choose $c$ to be the point where the expression inside the absolute value is zero — that's where the "kink" happens.

> [!FORMULA]
> **Split property for absolute values:**
> $$
> \int_a^b |x - k|\,dx = \int_a^k (k - x)\,dx + \int_k^b (x - k)\,dx
> $$
> This works because $|x-k| = k-x$ when $x \le k$, and $|x-k| = x-k$ when $x \ge k$.

Let's walk through it.

1. **Find the split point.**  
   Set $x-1 = 0 \implies x = 1$. This point lies inside $[0,4]$, so we break the integral at $x=1$:
   $$
   \int_{0}^{4}|x-1|\,dx = \int_{0}^{1}|x-1|\,dx + \int_{1}^{4}|x-1|\,dx
   $$

2. **Remove the absolute value on each piece.**  
   - On $[0,1]$, $x-1 \le 0$, so $|x-1| = -(x-1) = 1-x$.  
   - On $[1,4]$, $x-1 \ge 0$, so $|x-1| = x-1$.

Therefore:
   $$
   \int_{0}^{4}|x-1|\,dx = \int_{0}^{1}(1-x)\,dx + \int_{1}^{4}(x-1)\,dx
   $$

3. **Integrate each part.**  
   - First integral:
     $$
     \int_{0}^{1}(1-x)\,dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left(1 - \frac{1}{2}\right) - (0 - 0) = \frac{1}{2}
     $$
   - Second integral:
     $$
     \int_{1}^{4}(x-1)\,dx = \left[ \frac{x^2}{2} - x \right]_{1}^{4}
     $$
     Evaluate at $x=4$: $\frac{16}{2} - 4 = 8 - 4 = 4$ …

Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{4}|x-1|\,dx$

Concept understanding — Definite Integral Piecewise

Integrating a Piecewise Function

The additivity property

The method

Method: Integrating an absolute value — split where the inside changes sign

Steps

Common Mistakes

Q.By using the properties of definite integrals, evaluate the integral ∫04∣x−1∣ dx\int_{0}^{4}|x-1|\,dx∫04​∣x−1∣dx

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Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{4}|x-1|\,dx$

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