Question

Evaluate the integral: \[ I = \int_{-3}^{3} |2 - x| dx. \] 

• A. \( 12 \)
• B. \( 16 \)
• C. \( 13 \)
• D. \( 25 \)

Hint:

When integrating absolute value functions, always identify the points where the function changes sign and split the integral accordingly.

Answer 1

The Correct Option is C

Step 1: Identify the Breakpoint
The given function is \( |2 - x| \), which changes its definition when \( 2 - x = 0 \Rightarrow x = 2 \). Thus, we split the integral at \( x = 2 \). 
Step 2: Split the Integral into Two Parts
\[ I = \int_{-3}^{2} (2 - x) dx + \int_{2}^{3} (x - 2) dx. \] 
Step 3: Evaluate the First Integral
\[ \int_{-3}^{2} (2 - x) dx = \left[ 2x - \frac{x^2}{2} \right]_{-3}^{2}. \] Evaluating at \( x = 2 \): \[ 2(2) - \frac{2^2}{2} = 4 - 2 = 2. \] Evaluating at \( x = -3 \): \[ 2(-3) - \frac{(-3)^2}{2} = -6 - \frac{9}{2} = -\frac{21}{2}. \] Thus, \[ \left[ 2x - \frac{x^2}{2} \right]_{-3}^{2} = 2 - (-\frac{21}{2}) = 2 + \frac{21}{2} = \frac{25}{2}. \] 
Step 4: Evaluate the Second Integral
\[ \int_{2}^{3} (x - 2) dx = \left[ \frac{x^2}{2} - 2x \right]_{2}^{3}. \] Evaluating at \( x = 3 \): \[ \frac{3^2}{2} - 2(3) = \frac{9}{2} - 6 = -\frac{3}{2}. \] Evaluating at \( x = 2 \): \[ \frac{2^2}{2} - 2(2) = \frac{4}{2} - 4 = -2. \] Thus, \[ \left[ \frac{x^2}{2} - 2x \right]_{2}^{3} = -\frac{3}{2} - (-2) = -\frac{3}{2} + 2 = \frac{1}{2}. \] 
Step 5: Compute the Final Result
\[ I = \frac{25}{2} + \frac{1}{2} = \frac{26}{2} = 13. \] Thus, the final result is: \[ \boxed{13} \]

Answer 2

Step 1: Identify the Breakpoint 

The function is \( f(x) = |2 - x| \), which changes at \( x = 2 \) because \( 2 - x = 0 \Rightarrow x = 2 \). So we split the integral: \[ I = \int_{-3}^{3} |2 - x| \, dx = \int_{-3}^{2} (2 - x) \, dx + \int_{2}^{3} (x - 2) \, dx \]

Step 2: Evaluate the First Integral

\[ \int_{-3}^{2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{-3}^{2} \] At \( x = 2 \): \( 2(2) - \frac{2^2}{2} = 4 - 2 = 2 \)
At \( x = -3 \): \( 2(-3) - \frac{(-3)^2}{2} = -6 - \frac{9}{2} = -\frac{21}{2} \) So: \[ \int_{-3}^{2} (2 - x) \, dx = 2 - (-\frac{21}{2}) = \frac{25}{2} \]

Step 3: Evaluate the Second Integral

\[ \int_{2}^{3} (x - 2) \, dx = \left[ \frac{x^2}{2} - 2x \right]_{2}^{3} \] At \( x = 3 \): \( \frac{9}{2} - 6 = -\frac{3}{2} \)
At \( x = 2 \): \( \frac{4}{2} - 4 = -2 \) So: \[ \int_{2}^{3} (x - 2) \, dx = -\frac{3}{2} - (-2) = \frac{1}{2} \]

Step 4: Final Result

\[ I = \frac{25}{2} + \frac{1}{2} = \frac{26}{2} = \boxed{13} \]