3.7.5 Integration, SPM Practice (Question 13)

Question 13:

\begin{array}{l} Given that y = \frac{x^{2}}{2 x - 1}, show that \\ \frac{d y}{d x} = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} . Hence, evaluate \int_{- 2}^{2} \frac{x (x - 1)}{4 {(2 x - 1)}^{2}} d x . \end{array}

Solution:

\begin{array}{l} y = \frac{x^{2}}{2 x - 1} \\ \frac{d y}{d x} = \frac{(2 x - 1) (2 x) - x (2)}{{(2 x - 1)}^{2}} \\ = \frac{4 x^{2} - 2 x - 2 x^{2}}{{(2 x - 1)}^{2}} \\ = \frac{2 x^{2} - 2 x}{{(2 x - 1)}^{2}} \\ = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} (shown) \\ \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{8} \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{4} \int_{- 2}^{2} \frac{x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} [(\frac{2^{2}}{2 (2) - 1}) - (\frac{{(- 2)}^{2}}{2 (- 2) - 1})] \\ = \frac{1}{8} [(\frac{4}{3}) - (\frac{4}{- 5})] \\ = \frac{1}{8} (\frac{32}{15}) \\ = \frac{4}{15} \end{array}