Topic 9 of 9 · Form 4
Integration and kinematics
Reverse of differentiation: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + c (n ≠ −1); ∫eᵃˣ dx = eᵃˣ/a + c; ∫1/x dx = ln x + c; ∫sin ax dx = −cos ax/a + c; ∫cos ax dx = sin ax/a + c; ∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/[a(n + 1)] + c.
Notes for this topic 1 file, in order
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Integration
Reverse of differentiation: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + c (n ≠ −1); ∫eᵃˣ dx = eᵃˣ/a + c; ∫1/x dx = ln x + c; ∫sin ax dx = −cos ax/a + c; ∫cos ax dx = sin ax/a + c; ∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/[a(n + 1)] + c.
Definite integrals and areas
Area under a curve between two limits; area between a curve and a line.
Worked example: Find the area enclosed by y = 4x − x² and the x-axis.
Roots x = 0, 4 → ∫₀⁴(4x − x²) dx = [2x² − x³/3] = 32 − 64/3 = 32/3
Finding curves from gradients
Integrate dy/dx and use a given point to find c.
Kinematics
Displacement s, velocity v = ds/dt and acceleration a = dv/dt; reverse with integration.
Worked example: A particle has velocity v = 6t − 3t² m/s. Find the displacement from t = 0 to t = 2 and when it is instantaneously at rest.
s = ∫₀²(6t − 3t²) dt = [3t² − t³] = 12 − 8 = 4 m; v = 0 when t = 0 or t = 2
Check yourself
- Find ∫(2x + 1)³ dx.
- Evaluate ∫₀^(π/2) cos x dx.
- The gradient of a curve is 3x² − 2 and it passes through (1, 4). Find its equation.
Show answers
- (2x + 1)⁴/8 + c
- 1
- y = x³ − 2x + 5
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