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secondary 4 | A Maths
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Date Posted: 3 years ago
Views: 203
∫ᵏ₋₂ 3(x + 2)² dx = 64
[3(x + 2)²⁺¹ / (2 + 1)]ᵏ₋₂ = 64
[(x + 2)³]ᵏ₋₂ = 64
[(k + 2)³] - [(-2 + 2)³] = 64
(k + 2)³ = 64
k + 2 = ³√64
k + 2 = 4
k = 2
[3(x + 2)²⁺¹ / (2 + 1)]ᵏ₋₂ = 64
[(x + 2)³]ᵏ₋₂ = 64
[(k + 2)³] - [(-2 + 2)³] = 64
(k + 2)³ = 64
k + 2 = ³√64
k + 2 = 4
k = 2
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