Q.Sketch the graph of $y = |x+3|$ and evaluate $\int_{-6}^{0} |x+3| dx$.

The key idea is that $|x+3|$ is piecewise linear, with a corner at $x = -3$. The integral is the area under the V-shaped graph.

Step 1 – Graph shape:

$y = |x+3|$ is a V with vertex at $(-3, 0)$. For $x \ge -3$, the line is $y = x+3$; for $x \le -3$, it is $y = -(x+3)$.

Step 2 – Split the integral at the corner:

Step 3 – Evaluate each part:

Step 4 – Sum:

The absolute value function creates a V-shaped graph. For $y = |x+3|$, the "corner" occurs where the inside expression equals zero: $x+3 = 0$, so $x = -3$. This is the point where the function switches from decreasing to increasing.

To the left of $x = -3$, the expression $x+3$ is negative, so $|x+3| = -(x+3) = -x-3$. To the right of $x = -3$, $x+3$ is positive, so $|x+3| = x+3$. This split is the foundation for evaluating the integral.

1. Sketch the graph

The graph is a V shape with its vertex at $(-3, 0)$. For $x < -3$, the line has slope $-1$; for $x > -3$, the line has slope $+1$. The graph passes through $(-6, 3)$ and $(0, 3)$.

2. Identify the split point

The integrand $|x+3|$ is not differentiable at $x = -3$, but it is continuous everywhere. For integration, we split the interval $[-6, 0]$ at $x = -3$:

• On $[-6, -3]$: $x+3 \leq 0$, so $|x+3| = -(x+3) = -x-3$.
• On $[-3, 0]$: $x+3 \geq 0$, so $|x+3| = x+3$.

3. Write the integral as a sum

4. Evaluate the first integral

At $x = -3$: $-\frac{9}{2} + 9 = \frac{9}{2}$

At $x = -6$: $-\frac{36}{2} + 18 = -18 + 18 = 0$

So the value is $\frac{9}{2} - 0 = \frac{9}{2}$.

5. Evaluate the second integral

At $x = 0$: $0 + 0 = 0$

At $x = -3$: $\frac{9}{2} - 9 = -\frac{9}{2}$

So the value is $0 - \left(-\frac{9}{2}\right) = \frac{9}{2}$.

6. Add the two results

Method: Integrating an absolute-value (modulus) function

Use this to evaluate $\int |g(x)|\,dx$ — the graph of a modulus is a V (or a series of straight pieces), and you must integrate each straight piece separately.

Steps

Step 1: Find the corner(s) — where the inside is zero.

Solve $g(x) = 0$. At such a point the expression inside the modulus changes sign, so the formula for $|g(x)|$ changes. This is the split point.

Step 2: Write the piecewise form.

On the interval where $g(x) \ge 0$, use $|g(x)| = g(x)$; where $g(x) \le 0$, use $|g(x)| = -g(x)$.

Step 3: Split the integral at the corner and integrate each piece.

where $c$ is the corner. Each piece is positive.

Step 4 (check): use the triangle areas.

For a V-shaped graph each piece is a triangle; $\tfrac12 \times \text{base} \times \text{height}$ should match each integral.

Common Mistakes

Mistake 1: Integrating $x+3$ across the whole interval without splitting.

Why it's wrong: on $[-6,-3]$ the quantity $x+3$ is negative, so $|x+3| \ne x+3$ there; integrating $x+3$ straight from $-6$ to $0$ gives $0$, which is not the area. Correct approach: split at the corner $x=-3$, using $-(x+3)$ on the left piece and $(x+3)$ on the right.

Mistake 2: Not locating the corner at $x=-3$.

Why it's wrong: the modulus changes formula where the inside is zero, $x+3=0 \Rightarrow x=-3$; missing this split point makes the whole evaluation invalid. Correct approach: each piece is a triangle of base 3 and height 3, area $\tfrac92$, totalling $9$.