Q.By using the properties of definite integrals, evaluate the integral $\int_{2}^{8}|x-5|\,dx$

The key idea is that the absolute value function creates a piecewise expression, so we split the integral at the point where the expression inside changes sign.

Step 1: Find where $x-5 = 0$. This gives $x = 5$, which lies inside the interval $[2, 8]$.

Step 2: Write $|x-5|$ as a piecewise function:

Step 3: Split and evaluate the integral:

Compute each: …

The absolute value function $|x-5|$ is not a simple polynomial — it has a sharp corner at $x=5$. This means the function behaves differently on either side of that point. For $x < 5$, the expression $x-5$ is negative, so $|x-5| = -(x-5) = 5-x$. For $x \geq 5$, it is non-negative, so $|x-5| = x-5$.

Because the integrand changes its algebraic form at $x=5$, we cannot integrate it as a single expression from $2$ to $8$. Instead, we split the integral at the point where the definition changes. This is a standard technique: whenever you see an absolute value inside a definite integral, locate where the expression inside becomes zero, and break the integral there.

Here is the step-by-step solution:

1. Find the splitting point.

   Set $x-5 = 0$, which gives $x = 5$. This point lies inside the interval $[2, 8]$, so we must split the integral at $x=5$.

2. Write the piecewise definition of $|x-5|$.

3. Split the integral at $x=5$.

4. Evaluate the first integral.

At $x=5$: $5(5) - \frac{25}{2} = 25 - 12.5 = 12.5$

At $x=2$: $5(2) - \frac{4}{2} = 10 - 2 = 8$

Subtract: $12.5 - 8 = 4.5$

5. Evaluate the second integral.

At $x=8$: $\frac{64}{2} - 40 = 32 - 40 = -8$

At $x=5$: $\frac{25}{2} - 25 = 12.5 - 25 = -12.5$

Subtract: $(-8) - (-12.5) = -8 + 12.5 = 4.5$

6. Add the two results. …

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

$|expression|$ is a piecewise function, so break the interval at the point where the inside is zero and integrate each piece with the correct sign.

Steps

Step 1: Find the break point.

Solve $\text{inside}=0$. If that value lies inside the limits, it is the split point; if it lies outside, no split is needed.

Step 2: Write the piecewise definition.

$|x-k| = \begin{cases} -(x-k), & x<k \\ \;\;\,x-k, & x\ge k. \end{cases}$ …

Common Mistakes

Mistake 1: Forgetting to split at $x=5$.

Why it's wrong: $|x-5|$ is piecewise about $x=5$; integrating $x-5$ straight from $2$ to $8$ gives $0$, which is clearly wrong for a positive area. Correct approach: split into $\int_2^5(5-x)+\int_5^8(x-5)$.

Mistake 2: Using the wrong branch on each side. …