Area Under the Curve
Area bounded by curve and tangent
Grade 12
Question:
<p>The area of the region above the \(x\)-axis bounded by the curve \(y = \tan x,\ 0 \leq x \leq \dfrac{\pi}{2}\) and the tangent to the curve at \(x = \dfrac{\pi}{4}\) is</p>
<p>\(\dfrac{1}{2}\left(\log 2 - \dfrac{1}{2}\right)\)</p>
<p>\(\dfrac{1}{2}\left(\log 2 + \dfrac{1}{2}\right)\)</p>
<p>\(\dfrac{1}{2}(1 - \log 2)\)</p>
<p>\(\dfrac{1}{2}(1 + \log 2)\)</p>
Step-by-Step Solution
Key Concept: Find the tangent line equation at x = π/4, identify where it intersects the x-axis, then compute the area between the curve and tangent line using definite integration with careful attention to which function is above the other.
<p><strong>Step 1:</strong> Find the tangent line at x = π/4.</p><p>At x = π/4: y = tan(π/4) = 1 and dy/dx = sec²(π/4) = 2</p><p>Tangent line: y - 1 = 2(x - π/4), so y = 2x - π/2 + 1</p><p><strong>Step 2:</strong> Find where the tangent line meets the x-axis.</p><p>Set y = 0: 0 = 2x - π/2 + 1, giving x = π/4 - 1/2</p><p><strong>Step 3:</strong> Determine the bounded region.</p><p>The region is bounded by: the curve y = tan x from x = 0 to x = π/4, the tangent line from x = π/4 - 1/2 to x = π/4, and the x-axis.</p><p><strong>Step 4:</strong> Calculate the area.</p><p>Area = ∫₀^(π/4) tan x dx - (area of triangle formed by tangent line and x-axis)</p><p>∫₀^(π/4) tan x dx = [-ln(cos x)]₀^(π/4) = -ln(1/√2) - (-ln 1) = ln√2 = (1/2)ln 2</p><p>Triangle area with base = π/4 - 1/2 and height = 1 is (1/2)(π/4 - 1/2)(1) = π/8 - 1/4</p><p>∴ Answer: (1/2)ln 2 - (π/8 - 1/4) = <strong>1/2 + (1/2)ln 2 - π/8</strong></p>
Correct Answer: C