**(a)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.4792$ | B1 cao | [1] |

**(b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $=\frac{1}{2}\times\ln 2\times[2.1333+2(\text{their } 1.4792+1.0079)+0.6667]$ | B1 M1 | |
| $=\frac{\ln 2}{2}\times 7.7742...=2.694332406...=2.69$ (2 dp) | A1 | awrt $2.69$ |
| | | [3] |

**(c)(i)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{u=1+3e^{-x}\} \Rightarrow \frac{du}{dx}=-3e^{-x}$ or $\frac{dx}{du}=\frac{-1}{(u-1)}$ | B1 oe | |
| $\int\frac{4e^{-x}}{3\sqrt{(1+3e^{-x})}}dx=\left\{\pm\lambda\int\frac{1}{\sqrt{u}}du\right\}$ | M1 | |
| $=\frac{4}{9}\int\frac{1}{\sqrt{u}}du$ | A1 oe | |
| $=\frac{4}{9}\left(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)\{+c\}$ | dM1 | giving $\pm\beta u^{\frac{1}{2}}$ |
| $=-\frac{8}{9}u^{\frac{1}{2}}\{+c\}$ | A1 | |
| $=-\frac{8}{9}\sqrt{(1+3e^{-x})}\{+c\}$ | A1 | |
| | | [5] |

**(c)(ii)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(-\frac{8}{9}\left(\sqrt{1+3e^{0}}-\sqrt{1+3e^{-3\ln 2}}\right)\right)$ | M1 | Applying limits of $x=-3\ln 2$ and $x=0$ to an expression of the form $\pm A\sqrt{(1+3e^{-x})}$ and subtracts either way round. See notes. |
| $=-\frac{8}{9}(\sqrt{4}-\sqrt{25})$ | | |
| $=\frac{8}{3}$ | A1 | $\frac{8}{3}$ or awrt $2.67$ |
| | | [2] |
| | | [11] |

**Notes for (b)**

B1: Outside brackets $\frac{\ln 2}{2}\times\ln 2$ or $\frac{\ln 2}{2}$ or awrt $0.35$ or $\frac{\text{awrt } 0.69}{2}$.

Also allow $-\frac{1}{2}\times\ln 2$ or $-\frac{\ln 2}{2}$ or awrt $-0.35$ or $-\text{awrt}\frac{0.69}{2}$.

M1: For structure of trapezium rule $[...........]$

A1: anything that rounds to $2.69$

Note: It is possible to award: (a) B0 (b) B1M1A1 (awrt 2.69)

Note: Working must be seen to demonstrate the use of the trapezium rule. Note: actual area is $2.66666...$

Note: Award B1M1A1 for $\frac{\ln 2}{2}(2.1333+0.6667)+\ln 2(\text{their } 1.4792+1.0079)=2.694332406...$

**Notes for (c)(i)**

B1: For $\frac{du}{dx}=-3e^{-x}$ or $du=-3e^{-x}dx$ or $\frac{dx}{du}=\frac{-1}{(u-1)}$ or $\frac{dx}{-3e^{-x}}=du$ or equivalent.

M1: Applying the substitution and achieving $\pm\lambda\int\frac{1}{\sqrt{u}}(du)$ or $\pm\lambda\int u^{-\frac{1}{2}}(du)$, $\lambda \neq 0$

Note: Any $(u-1)$ terms need to be cancelled out for this M1 mark.

A1: $\frac{4}{9}\int\frac{1}{\sqrt{u}}du$ or $-\frac{4}{9}\int u^{-\frac{1}{2}}(du)$ or equivalent.

dM1: (dependent on the first M mark) Integrates $\pm\lambda\int\frac{1}{\sqrt{u}}du$ to give $\pm\beta u^{\frac{1}{2}}, \lambda \neq 0, \beta \neq 0$

A1: $-\frac{8}{9}\sqrt{(1+3e^{-x})}, \text{simplified or un-simplified, with/without } +c$

Note: $\int\frac{4(du)}{3\sqrt{u}} \times \frac{-du}{(u-1)}$ is M0A0 unless the $(u-1)$ terms have been cancelled out later

but $\int\frac{4(u-1)}{3\sqrt{u}} \times \frac{-du}{(u-1)}$ is M1A1.

**Notes for (c)(ii)**

M1: Applies limits of $x=-3\ln 2$ or $-2.07...$ and $x=0$ to an expression of the form $\pm A\sqrt{(1+3e^{-x})}$ and subtracts either way round.

Or attempts to apply limits of $u=25$ and $u=4$ to an expression in the form $\pm\beta u^{\frac{1}{2}}$ and subtracts either way round.

A1: $\frac{8}{3}$ or anything that rounds to $2.67$.

Note: The final A1 mark in (c)(ii) is dependent on (c)(i) B1M1A1M1 and (c)(ii) M1.

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