2026 NORTHBROOKS P1 Q6

Given that \(y=x^2\sqrt{1-3x}\), show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{x(4-15x)}{2\sqrt{1-3x}}\).[3]

Hence find the value of \(\displaystyle\int_{-1}^{0}\dfrac{2+x(4-15x)}{2\sqrt{1-3x}}\,\mathrm{d}x\).[3]