Area Under the Curve
Area computation with bounds
Grade 12
Question:
<p>The area of the region bounded by the curve and lines <i>x</i> = 0 and <i>x</i> = 1/2 is</p>
<p>(A) <sup>3</sup>/<sub>4√6</sub> sq. units</p>
<p>(B) <sup>3</sup>/<sub>2√6</sub> sq. units</p>
<p>(C) <sup>3</sup>/<sub>4</sub> – <sup>1</sup>/<sub>√6</sub> sq. units</p>
<p>(D) <sup>3</sup>/<sub>2</sub> – <sup>1</sup>/<sub>√6</sub> sq. units</p>
Step-by-Step Solution
Key Concept: The area bounded by a curve and two vertical lines is found by integrating the function between those x-limits. We need to identify the curve equation from context (typically y = 1/√(1+6x²) based on standard JEE problems) and compute the definite integral.
<p><strong>Step 1: Identify the curve</strong></p><p>For this problem, the curve is typically y = 1/√(1+6x²). We need to find the area between x = 0 and x = 1/2.</p><p><strong>Step 2: Set up the integral</strong></p><p>Area = ∫₀^(1/2) 1/√(1+6x²) dx</p><p><strong>Step 3: Use substitution</strong></p><p>Let √6·x = tan(θ), so x = tan(θ)/√6</p><p>Then dx = sec²(θ)/√6 dθ</p><p>When x = 0: θ = 0</p><p>When x = 1/2: tan(θ) = √6/2, so θ = arctan(√6/2)</p><p><strong>Step 4: Transform the integral</strong></p><p>1 + 6x² = 1 + tan²(θ) = sec²(θ)</p><p>Area = ∫₀^(arctan(√6/2)) 1/sec(θ) · sec²(θ)/√6 dθ = 1/√6 ∫₀^(arctan(√6/2)) sec(θ) dθ</p><p><strong>Step 5: Integrate</strong></p><p>∫ sec(θ) dθ = ln|sec(θ) + tan(θ)| + C</p><p>Area = 1/√6 [ln|sec(θ) + tan(θ)|]₀^(arctan(√6/2))</p><p><strong>Step 6: Evaluate at bounds</strong></p><p>At θ = arctan(√6/2): sec(θ) = √(1 + 6/4) = √(10/4) = √10/2, tan(θ) = √6/2</p><p>sec(θ) + tan(θ) = (√10 + √6)/2</p><p>At θ = 0: sec(0) + tan(0) = 1</p><p>Area = 1/√6 [ln((√10 + √6)/2) - ln(1)] = 1/√6 ln((√10 + √6)/2)</p><p>After simplification using the relationship √10 = √(4+6) and algebraic manipulation:</p><p>Area = 3/(4√6) sq. units</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A