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

The key idea is that $|x+2|$ is not differentiable at $x=-2$, so we split the integral at that point.

Step 1: Write the absolute value as a piecewise function:

Step 2: Split the integral at $x=-2$:

Step 3: Evaluate each part:

$\int_{-5}^{-2} -(x+2)\,dx = -\left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} = -\left( (2 - 4) - \left(\frac{25}{2} - 10\right) \right) = -\left( -2 - \frac{5}{2} \right) = \frac{9}{2}$ …

The absolute value function $|x+2|$ creates a sharp corner at $x = -2$, where the expression inside changes sign. To integrate, we must remove the absolute value by considering the piecewise definition:

This is the core idea: break the integral at the point where the expression inside the absolute value equals zero. The given interval $[-5, 5]$ spans both sides of $x = -2$, so we split the integral into two parts.

1. Identify the split point. Solve $x+2 = 0 \Rightarrow x = -2$. This lies inside $[-5, 5]$, so we write:

2. Evaluate the left part ($x$ from $-5$ to $-2$). Here $x < -2$, so $x+2 < 0$, meaning $|x+2| = -(x+2) = -x - 2$.

Compute the antiderivative: $\int (-x - 2) \, dx = -\frac{x^2}{2} - 2x$.

Apply the limits:

Simplify term by term:

• At $x = -2$: $-\frac{4}{2} + 4 = -2 + 4 = 2$
• At $x = -5$: $-\frac{25}{2} + 10 = -12.5 + 10 = -2.5$

So the difference is $2 - (-2.5) = 4.5 = \frac{9}{2}$.

3. Evaluate the right part ($x$ from $-2$ to $5$). Here $x \geq -2$, so $x+2 \geq 0$, meaning $|x+2| = x+2$.

Antiderivative: $\int (x+2) \, dx = \frac{x^2}{2} + 2x$.

Apply limits:

Simplify:

• At $x = 5$: $12.5 + 10 = 22.5 = \frac{45}{2}$
• At $x = -2$: $2 - 4 = -2$ …

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: Splitting the interval at $x=0$ instead of where $x+2=0$.

Why it's wrong: $|x+2|$ changes form at $x=-2$, not $x=0$; splitting at the wrong point gives a wrong sign on part of the range. Correct approach: solve $x+2=0\Rightarrow x=-2$ and break there.

Mistake 2: Integrating $x+2$ over the whole $[-5,5]$ without the sign flip. …