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junior college 1 | H2 Maths
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Date Posted: 2 years ago
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When y = 0, t = -3, x = 20
When x = 0, t = 1 or t = 2
We use t = 1 for x = 0 since that is the point where the curve and y axis forms the bound with the x-axis and the x-intercept.
To find the entire area bound by both axes and the curve, integrate y with respect to t
Area = ∫₀²⁰ y dx
Normally for typical y = f(x) curves, you'll integrate y with respect to x but here it is almost impossible to express y in terms of x. So that's why the curve is described by a third variable, t)
We also need to change the bound into terms of t and also the dx into in terms of dt.
x = 2 - 3t + t², → dx/dt = 2t - 3 → "dx = (2t - 3) dt"
So, Area = ∫₀²⁰ y dx
= ∫₁⁻³ [(3 + t)² (2t - 3)] dt
= ∫₁⁻³ [(9 + 6t + t²)(2t - 3)] dt
= ∫₁⁻³ (2t³ + 9t² - 27) dt
= [½t⁴ + 3t³ - 27t]₁⁻³
[½(-3)⁴ + 3(-3)³ - 27(-3)] - [½(1) + 3(1³) - 27(1)]
= 40½ + 23½
= 64
When x = 0, t = 1 or t = 2
We use t = 1 for x = 0 since that is the point where the curve and y axis forms the bound with the x-axis and the x-intercept.
To find the entire area bound by both axes and the curve, integrate y with respect to t
Area = ∫₀²⁰ y dx
Normally for typical y = f(x) curves, you'll integrate y with respect to x but here it is almost impossible to express y in terms of x. So that's why the curve is described by a third variable, t)
We also need to change the bound into terms of t and also the dx into in terms of dt.
x = 2 - 3t + t², → dx/dt = 2t - 3 → "dx = (2t - 3) dt"
So, Area = ∫₀²⁰ y dx
= ∫₁⁻³ [(3 + t)² (2t - 3)] dt
= ∫₁⁻³ [(9 + 6t + t²)(2t - 3)] dt
= ∫₁⁻³ (2t³ + 9t² - 27) dt
= [½t⁴ + 3t³ - 27t]₁⁻³
[½(-3)⁴ + 3(-3)³ - 27(-3)] - [½(1) + 3(1³) - 27(1)]
= 40½ + 23½
= 64
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