Sketch the graph of and evaluate .
The value of the integral is 9 square units.
Explanation
This question tests understanding of absolute value functions and definite integrals. The function y = |x + 3| has a vertex at x = -3 where it changes definition. The integral must be split at this point. From the context, when a curve is above the x-axis, the area is given by the integral directly. Both portions of this V-shaped graph lie above the x-axis, so we evaluate the integral by splitting it at the vertex.
Solution Steps
Step 1: Sketch the graph of y = |x + 3| The function y = |x + 3| is a V-shaped graph with vertex at (-3, 0).
- For x -3: y = x + 3 (line with slope 1)
- For x < -3: y = -(x + 3) = -x - 3 (line with slope -1) The graph passes through points (-6, 3), (-3, 0), and (0, 3).
Step 2: Split the integral at the vertex Since the definition changes at x = -3, we split the integral: ∫₋₆⁰ |x + 3| dx = ∫₋₆⁻³ (-x - 3) dx + ∫₋₃⁰ (x + 3) dx
Step 3: Evaluate the first integral (from -6 to -3) ∫₋₆⁻³ (-x - 3) dx = [-x²/2 - 3x]₋₆⁻³ = [-(−3)²/2 - 3(−3)] - [-(−6)²/2 - 3(−6)] = [-9/2 + 9] - [-18 + 18] = 9/2 - 0 = 9/2
Step 4: Evaluate the second integral (from -3 to 0) ∫₋₃⁰ (x + 3) dx = [x²/2 + 3x]₋₃⁰ = [0 + 0] - [(−3)²/2 + 3(−3)] = 0 - [9/2 - 9] = 0 - (-9/2) = 9/2
Step 5: Add both areas Total Area = 9/2 + 9/2 = 9 square units