Question

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

• A. 14
• B. 16
• C. 8
• D. 10
• E. 12

Hint:

If the function does not change sign in the interval, you can simply remove the absolute value bars.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to evaluate a definite integral of an absolute value function. The key is to analyze the sign of the expression inside the absolute value over the interval of integration.
Step 2: Key Formula or Approach:
The absolute value function is defined as:
\[ |f(x)| = \begin{cases} f(x) & \text{if } f(x) \geq 0
-f(x) & \text{if } f(x)<0 \end{cases} \] We need to find where the argument \( x+3 \) changes sign. This occurs at \( x+3=0 \), i.e., \( x=-3 \).
Then we examine the interval of integration \( [-2, 2] \) in relation to this point.
Step 3: Detailed Explanation:
The expression inside the absolute value is \( x+3 \).
The sign change occurs at \( x=-3 \).
The interval of integration is from -2 to 2.
Let's check the sign of \( x+3 \) within this interval. For any \( x \) in \( [-2, 2] \), the smallest value is \( x=-2 \).
At \( x=-2 \), \( x+3 = -2+3 = 1 \), which is positive.
Since \( x+3 \) is positive at the start of the interval and is an increasing function, it will be positive for the entire interval \( [-2, 2] \).
Therefore, for \( x \in [-2, 2] \), we have \( |x+3| = x+3 \).
So, the integral simplifies to:
\[ \int_{-2}^2 (x+3) dx \] Now we evaluate this standard definite integral:
\[ \left[ \frac{x^2}{2} + 3x \right]_{-2}^2 \] \[ = \left( \frac{2^2}{2} + 3(2) \right) - \left( \frac{(-2)^2}{2} + 3(-2) \right) \] \[ = \left( \frac{4}{2} + 6 \right) - \left( \frac{4}{2} - 6 \right) \] \[ = (2 + 6) - (2 - 6) \] \[ = 8 - (-4) = 8 + 4 = 12 \] Step 4: Final Answer:
The value of the integral is 12.