2019 H2 Mathematics Paper 2 Question 1

Integration Techniques

Answers

{-\frac{2}{3} x \left( 1 - x \right)^{\frac{3}{2}}}\allowbreak {- \frac{4}{15} \left(1 - x\right)^{\frac{5}{2}} + C}

{-\frac{2}{3} (1-x)^{\frac{3}{2}}}\allowbreak {+ \frac{2}{5} (1-x)^{\frac{5}{2}} + C'}

\begin{align*} & \int x \left(1 - x\right)^{\frac{1}{2}} \; \mathrm{d}x \\ & = x \left( - \frac{2}{3} \left(1 - x\right)^{\frac{3}{2}} \right) - \int - \frac{2}{3} \left(1 - x\right)^{\frac{3}{2}} \; \mathrm{d}x \\ & = -\frac{2}{3} x \left( 1 - x \right)^{\frac{3}{2}} - \frac{4}{15} \left(1 - x\right)^{\frac{5}{2}} + C \; \blacksquare \end{align*}

2u \frac{\mathrm{d}u}{\mathrm{d}x} = -1

\begin{align*} & \int x \left(1 - x\right)^{\frac{1}{2}} \; \mathrm{d}x \\ & = \int (1-u^2) u (-2u) \; \mathrm{d} u \\ & = \int - 2 u^2 + 2 u^4 \; \mathrm{d}u \\ & = - \frac{2}{3} u^3 + \frac{2}{5} u^5 + C' \\ & = -\frac{2}{3} (1-x)^{\frac{3}{2}} + \frac{2}{5} (1-x)^{\frac{5}{2}} + C' \; \blacksquare \end{align*}

Difference between the two answers

\begin{align*} &= -\frac{2}{3} (1-x)^{\frac{3}{2}} + \frac{2}{5} (1-x)^{\frac{5}{2}} + C' \\ &\qquad \quad - \left( -\frac{2}{3} x \left( 1 - x \right)^{\frac{3}{2}} - \frac{4}{15} \left(1 - x\right)^{\frac{5}{2}} + C \right) \\ &= \left( -\frac{2}{3} + \frac{2}{3}x \right) (1 - x)^{\frac{3}{2}} + \left( \frac{2}{5} + \frac{4}{15} \right) (1-x)^{\frac{5}{2}} + C' - C \\ &= -\frac{2}{3} \left( 1-x \right) (1 - x)^{\frac{3}{2}} + \frac{10}{15} (1-x)^{\frac{5}{2}} + C' - C \\ &= -\frac{2}{3} (1 - x)^{\frac{5}{2}} + \frac{2}{3} (1-x)^{\frac{5}{2}} + C' - C \\ &= C' - C \end{align*}

Hence the two answers differ by a constant

{\blacksquare.}