17M.1.SL.TZ2.S_5

pestleMathematicsAISLPaper 117M· ahl-5-11-indefinite-integration-reverse-chain-by-substitutionsource ↗

Let $f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}$ . Given that $f(0) = 1$ , find $f(x)$ .

Markscheme / solution

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

valid approach (M1)

eg $\,\,\,\,\,$ $\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} }$

correct integration by substitution/inspection A2

eg $\,\,\,\,\,$ $f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + c,{\text{ }}\frac{{ - 1}}{{4{{({x^3} + 1)}^4}}}$

correct substitution into their integrated function (must include $c$ ) M1

eg $\,\,\,\,\,$ $1 = \frac{{ - 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} - \frac{1}{4} + c = 1$

Note: Award M0 if candidates substitute into $f^{'}$ or $f^{″}$ .

$c = \frac{5}{4}$ (A1)

$f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ - 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} - 1}}{{4{{({x^3} + 1)}^4}}}} \right)$ A1 N4

[6 marks]

Examiners’ report

[N/A]