2018 H2 Mathematics Paper 1 Question 1

Integration Techniques

Answers

(i)

dydx=1x2−ln⁡xx2{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x^2} - \frac{\ln x}{x^2}}

(ii)

1−2e{1 - \frac{2}{\mathrm{e}}}

Full solutions

(i)

dydx=1x2−ln⁡xx2  ■\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x^2} - \frac{\ln x}{x^2} \; \blacksquare

(ii)

dydx=1x2−ln⁡xx2y=∫(1x2−ln⁡xx2)  dx=∫1x2  dx−∫ln⁡xx2  dx=−1x−∫ln⁡xx2  dx∫ln⁡xx2  dx=−1x−y+C∫ln⁡xx2  dx=−1x−ln⁡xx+C\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{x^2} - \frac{\ln x}{x^2} \\ y &= \int \left( \frac{1}{x^2} - \frac{\ln x}{x^2} \right) \; \mathrm{d}x \\ &= \int \frac{1}{x^2} \; \mathrm{d}x - \int \frac{\ln x}{x^2} \; \mathrm{d}x \\ &= - \frac{1}{x} - \int \frac{\ln x}{x^2} \; \mathrm{d}x \\ \int \frac{\ln x}{x^2} \; \mathrm{d}x &= - \frac{1}{x} - y + C \\ \int \frac{\ln x}{x^2} \; \mathrm{d}x &= - \frac{1}{x} - \frac{\ln x}{x} + C \\ \end{align*}
∫1eln⁡xx2  dx=[−1x−ln⁡xx]1e=−1e−ln⁡ee+1−0=1−2e  ■\begin{align*} & \int_1^\mathrm{e} \frac{\ln x}{x^2} \; \mathrm{d}x \\ &= \left[ - \frac{1}{x} - \frac{\ln x}{x} \right]_1^\mathrm{e} \\ &= -\frac{1}{\mathrm{e}} - \frac{\ln \mathrm{e}}{\mathrm{e}} + 1 - 0 \\ &= 1 - \frac{2}{\mathrm{e}} \; \blacksquare \end{align*}