SPM.1.SL.TZ0.4

pestleMathematicsAASLPaper 1SPM· sl-5-10-indefinite-integration-reverse-chain-by-substitutionsource ↗

Let $f'\left( x \right) = \frac{{8x}}{{\sqrt {2{x^2} + 1} }}$ . Given that $f\left( 0 \right) = 5$ , find $f\left( x \right)$ .

Markscheme / solution

attempt to integrate (M1)

$u = 2{x^2} + 1 \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = 4x$

$\int {\frac{{8x}}{{\sqrt {2{x^2} + 1} }}} {\text{d}}x = \int {\frac{2}{{\sqrt u }}} {\text{d}}u$ (A1)

EITHER

$= 4\sqrt u \left( { + C} \right)$ A1

$= 4\sqrt {2{x^2} + 1} \left( { + C} \right)$ A1

THEN

correct substitution into their integrated function (must have C) (M1)

$5 = 4 + C \Rightarrow C = 1$

$f\left( x \right) = 4\sqrt {2{x^2} + 1} + 1$ A1

[5 marks]

Examiners’ report

[N/A]