Area Under the Curve
Area between two curves
Grade 12
Question:
<p>The region bounded by the curves \(y = 2^x\) and \(y = |x + 1|\) has area equal to:</p>
<p>\(\dfrac{3}{2} - \dfrac{2}{\log_e 2} + \dfrac{1}{\log_e 2}\)</p>
<p>\(\dfrac{3}{2} - \dfrac{1}{\log_e 2}\)</p>
<p>\(\dfrac{3}{2} + \dfrac{1}{\log_e 2}\)</p>
<p>\(\dfrac{3}{2} + \dfrac{2}{\log_e 2}\)</p>
Step-by-Step Solution
Key Concept: Find intersection points of y = 2^x and y = |x+1| by considering cases for the absolute value, then integrate the difference of functions over the appropriate interval.
<p><strong>Step 1:</strong> Identify the absolute value branches: y = |x+1| means y = x+1 for x ≥ -1 and y = -(x+1) for x < -1.</p><p><strong>Step 2:</strong> Find intersections with y = 2^x.</p><p>For x ≥ -1: Solve 2^x = x + 1. By inspection, x = 0 gives 2^0 = 1 = 0+1 ✓</p><p>For x < -1: Solve 2^x = -(x+1). At x = -1: 2^(-1) = 1/2 and 0 ✗. As x → -∞, 2^x → 0 and -(x+1) → ∞, so they must intersect. Testing: at x = -2, 2^(-2) = 1/4 and -(-2+1) = 1, so intersection exists between -2 and -1. Solving 2^x = -x - 1 numerically gives x ≈ -1.256.</p><p><strong>Step 3:</strong> Determine which function is above the other in each region and set up the integral.</p><p>For x ∈ [-1.256, 0]: y = x+1 is above y = 2^x</p><p><strong>Step 4:</strong> Calculate the area:</p><p>Area = ∫₍₋₁.₂₅₆₎⁰ [(x+1) - 2^x] dx</p><p>= [x²/2 + x - 2^x/ln(2)]₍₋₁.₂₅₆₎⁰</p><p>= [0 + 0 - 1/ln(2)] - [(a²/2 + a - 2^a/ln(2))] where a ≈ -1.256</p><p>∴ Answer: B</p>
Correct Answer: B