Question

$\int_{-3}^{3}|x+2|dx=$

• A. 17
• B. 9
• C. 14
• D. 13
• E. 12

Hint:

Geometry Tip: You can also solve this using the area of triangles! Draw the graph of $y = |x+2|$. It forms a small triangle on the left (base 1, height 1, Area = 0.5) and a large triangle on the right (base 5, height 5, Area = 12.5). Total area = 13.

Answer 1

The Correct Option is D

Concept:
To integrate an absolute value function, you must split the integral at its critical point (where the expression inside the absolute value equals zero). For $|x+2|$, the critical point is $x = -2$. The function behaves as $-(x+2)$ when $x < -2$, and as $(x+2)$ when $x \ge -2$.

Step 1: Identify the critical point and split the integral.
Set the inside of the absolute value to zero: $x + 2 = 0 \implies x = -2$. Split the integral from $-3$ to $3$ at this point: $$I = \int_{-3}^{-2} -(x+2) dx + \int_{-2}^{3} (x+2) dx$$

Step 2: Integrate the first piece.
Evaluate the integral for the negative portion: $$I_1 = \int_{-3}^{-2} (-x - 2) dx = \left[ -\frac{x^2}{2} - 2x \right]_{-3}^{-2}$$ Substitute the limits: $$I_1 = \left( -\frac{(-2)^2}{2} - 2(-2) \right) - \left( -\frac{(-3)^2}{2} - 2(-3) \right)$$ $$I_1 = (-2 + 4) - (-4.5 + 6) = 2 - 1.5 = 0.5$$

Step 3: Integrate the second piece.
Evaluate the integral for the positive portion: $$I_2 = \int_{-2}^{3} (x + 2) dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{3}$$ Substitute the limits: $$I_2 = \left( \frac{(3)^2}{2} + 2(3) \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right)$$ $$I_2 = (4.5 + 6) - (2 - 4) = 10.5 - (-2) = 12.5$$

Step 4: Sum the two areas.
Add the results of the two definite integrals: $$I = I_1 + I_2 = 0.5 + 12.5 = 13$$ Hence the correct answer is (D) 13.