Question

The area bounded by the lines \( y = 1 + |x + 1| \), \( x = -3 \), \( x = 3 \) and \( y = 0 \) is

• A. \( 14 \text{ Square units} \)
• B. \( 15 \text{ Square units} \)
• C. \( 16 \text{ Square units} \)
• D. \( 17 \text{ Square units} \)

Hint:

For absolute value functions, always split the interval at the point where the expression inside the modulus is zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The area can be calculated using the definite integral of the function \( f(x) = 1 + |x + 1| \) from \( x = -3 \) to \( x = 3 \).

Step 2: Key Formula or Approach:
The integral is \( A = \int_{-3}^{3} (1 + |x + 1|) dx \).
Split the integral at \( x = -1 \) where the absolute value changes sign.

Step 3: Detailed Explanation:
\( A = \int_{-3}^{-1} (1 - (x + 1)) dx + \int_{-1}^{3} (1 + (x + 1)) dx \)
\( A = \int_{-3}^{-1} (-x) dx + \int_{-1}^{3} (x + 2) dx \)
\( A = [-\frac{x^2}{2}]_{-3}^{-1} + [\frac{x^2}{2} + 2x]_{-1}^{3} \)
\( A = (-\frac{1}{2} - (-\frac{9}{2})) + ((\frac{9}{2} + 6) - (\frac{1}{2} - 2)) \)
\( A = (4) + (10.5 - (-1.5)) = 4 + 12 = 16 \).

Step 4: Final Answer:
The total area is \( 16 \) square units.