1 Integration Q

33(a)(i) As A1 is trapeziumA1 = (α^2 + β^2)(β - α)/2(ii) (β^3 - α^3)/3(iii) A2 = (α^2 + β^2)(β - α)/2 - (β^3 - α^3)/3= (3β^3 - 3αβ^2 + 3α^2β - 3α^3 + 2α^3 - 2β^3)/6= (β^3 - 3αβ^2 + 3α^2β - α^3 )/6= (β-α)^3/6(b)(i) Sub. y = x^2 into y = mx + kx^2 - mx - k = 0 and the roots are α and βNow, as (2,5) is on y = mx + k => k = 5 - 2mSo, x^2 - mx + 2m - 5 = 0 and the roots are α and β(ii) (β-α)^2 = (α + β)^2 - 4αβ= m^2 - 4(2m - 5)= m^2 - 8m + 20If (β-α)^2 is minimum, (β-α)^3 is also minimumNow as k = 5 - 2m and k >= 0 => 0 < m <= 5/2 Consider f(m) = m^2 - 8m + 20. f'(m) = 2m - 8 < 0 for the possible range of m So, when m attains its maximum value 5/2, A2 = (β-α)^3/6 = 125/48 is minimum

(b)(ii)
β>α⇒(β-α)=√(β-α)²=√[(β+α)²-4αβ]=√[m²-4(2m-5)]=√[(m-4)²+4]
A₂=(β-α)³/6=√[(m-4)²+4]³/6
A₂is min when m=4 and min value of A₂is √(4³)/6=4/3