TMUA 2017 Paper 1 Question 1 — Worked Solution

TMUA 2017 · Paper 1 Differentiation and integrationIntegration · finding the constant from a point6 options

Given that dydx = 3x² − 2 − 3xx³, x ≠ 0 and y = 5 when x = 1, find y in terms of x.

Ay = 13x³ + x⁻² − 3x⁻¹ + 623
By = x³ + 12x⁻² − 3x⁻¹ + 612
Cy = x³ + x⁻² − 3x⁻¹ + 6
Dy = x³ + x⁻² − x⁻¹ + 4
Ey = x³ + 2x⁻² − x⁻¹ + 3
Fy = 3x³ + x⁻² − x⁻¹ + 2

Show the answer and worked solution

answer · C

Ay = 13x³ + x⁻² − 3x⁻¹ + 623
By = x³ + 12x⁻² − 3x⁻¹ + 612
Cy = x³ + x⁻² − 3x⁻¹ + 6
Dy = x³ + x⁻² − x⁻¹ + 4
Ey = x³ + 2x⁻² − x⁻¹ + 3
Fy = 3x³ + x⁻² − x⁻¹ + 2

Split the fraction into powers before integrating: 2−3xx³ = 2x⁻³ − 3x⁻², so dydx = 3x² − 2x⁻³ + 3x⁻². Integrating term by term gives x³ − 2⋅x⁻²−2 + 3⋅x⁻¹−1 + c = x³ + x⁻² − 3x⁻¹ + c. Putting x = 1, y = 5 gives 1 + 1 − 3 + c = 5, so c = 6. The tempting slip is to forget that dividing by x³ turns the −3x into −3x⁻², which integrates to +3x⁻¹ before the sign of the term is applied.