16N.1.SL.TZ0.S_6

pestleMathematicsAISLPaper 116N· ahl-5-11-indefinite-integration-reverse-chain-by-substitutionsource ↗

Let $f'(x) = {\sin ^3}(2x)\cos (2x)$ . Find $f(x)$ , given that $f\left( {\frac{\pi }{4}} \right) = 1$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of integration (M1)

eg $\,\,\,\,\,$ $\int {f'(x){\text{d}}x}$

correct integration (accept missing $C$ ) (A2)

eg $\,\,\,\,\,$ $\frac{1}{2} \times \frac{{{{\sin }^4}(2x)}}{4},{\text{ }}\frac{1}{8}{\sin ^4}(2x) + C$

substituting initial condition into their integrated expression (must have $+ C$ ) M1

eg $\,\,\,\,\,$ $1 = \frac{1}{8}{\sin ^4}\left( {\frac{\pi }{2}} \right) + C$

Note: Award M0 if they substitute into the original or differentiated function.

recognizing $\sin \left( {\frac{\pi }{2}} \right) = 1$ (A1)

eg $\,\,\,\,\,$ $1 = \frac{1}{8}{(1)^4} + C$

$C = \frac{7}{8}$ (A1)

$f(x) = \frac{1}{8}{\sin ^4}(2x) + \frac{7}{8}$ A1 N5

[7 marks]

Examiners’ report

[N/A]