Integration: Area Under a Curve

1. 1 If f'(x) = 3x², what is f(x)? Reverse the power rule: add 1 to the power, divide by the new power. x³. But also x³+5, x³−7... 'We need a constant: f(x) = x³ + c.' This is the indefinite integral. ∫3x² dx = x³ + c.
2. 2 ∫xⁿ dx = xⁿ⁺¹/(n+1) + c (for n ≠ −1). Practise: ∫4x dx = 2x² + c. ∫x³ dx = x⁴/4 + c. ∫(3x²+2x−1) dx = x³+x²−x+c. Students integrate six polynomial expressions.
3. 3 The definite integral ∫[a to b] f(x) dx evaluates between limits: compute the integral, substitute b, subtract the result with a. ∫[1 to 3] x² dx = [x³/3] from 1 to 3 = 27/3 − 1/3 = 8⅔. No constant c needed for definite integrals.
4. 4 ∫[a to b] f(x) dx gives the signed area between y=f(x) and the x-axis from a to b. Positive when curve is above x-axis. Sketch y=x²−4. Show area from x=2 to x=4. Calculate: ∫[2 to 4] (x²−4) dx = [x³/3−4x] from 2 to 4 = (64/3−16)−(8/3−8) = 56/3−8 = 32/3 ≈ 10.7.
5. 5 For y=x²−4, find area from x=0 to x=2 (the region below the x-axis). The integral gives −8/3 (negative). The actual area is 8/3. 'Always sketch first to identify regions above and below the axis. For area, take the modulus of negative integrals.'
6. 6 Area between y=x+4 and y=x² from their intersections. Intersections: x²=x+4 → x²−x−4=0 → x=−1.56, 2.56 (approx). Area = ∫[−1.56 to 2.56] (x+4−x²) dx. Integrate (top minus bottom curve).
7. 7 Show that definite integration gives exact area while the trapezium rule approximates it. For simple cases, compute both and compare. Discuss when the trapezium rule is needed (non-polynomial functions).

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