Application of Definite Integration

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Area A = ∫[a to b] f(x) dx This represents the area of the region between the curve y = f(x) and the X-axis from x = a to x = b. If f(x) ≥ 0 in the interval [a,b], the area is positive. If f(x) < 0, …

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Step 1: Set up the integral A = ∫[1 to 3] x² dx Step 2: Integrate A = [x³/3]₁³ Step 3: Evaluate A = (3³/3) - (1³/3) A = 27/3 - 1/3 A = 26/3 square units Answer: 26/3 square units…

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Standard form: y² = 4ax (opens rightward) To find area bounded by this parabola and line x = k: 1. Solve for y: y = ±2√(ax) 2. Due to symmetry: A = 2∫[0 to k] 2√(ax) dx 3. Integrate: A = 4√a × (2/3) …

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When f(x) < 0 in interval [a,b]: 1. The integral ∫[a to b] f(x) dx gives a negative value 2. For actual area, take absolute value: |∫[a to b] f(x) dx| 3. If curve crosses X-axis at point x = c, split…

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Step 1: Identify where curve crosses X-axis y = -2x = 0 when x = 0 Step 2: Split into two regions A₁ (x = -1 to x = 0): curve above X-axis A₂ (x = 0 to x = 2): curve below X-axis Step 3: Calculate A…

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Area A = ∫[c to d] g(y) dy Used when: - Curve is given as x = g(y) - Region is bounded by Y-axis and lines y = c, y = d - It's easier to integrate with respect to y Example: For curve x² = 16y betwe…

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Step 1: Express x in terms of y x² = 16y → x = 4√y (first quadrant, x ≥ 0) Step 2: Set up integral with respect to y A = ∫[1 to 4] x dy = ∫[1 to 4] 4√y dy Step 3: Integrate A = 4∫[1 to 4] y^(1/2) dy…

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Ellipse: x²/a² + y²/b² = 1 Area = πab square units Derivation: Step 1: Solve for y: y = ±(b/a)√(a² - x²) Step 2: Use symmetry: A = 4∫[0 to a] (b/a)√(a² - x²) dx Step 3: Apply standard integral: ∫√(a²…