# Question 2:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x=1.4$, $y=0.6595$ (4 dp) | B1 cao | 0.6595 — look for this on table or in candidate's working |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\times(0.2)$ outside brackets | B1 o.e. | Outside brackets $\frac{1}{2}\times(0.2)$ or $\frac{1}{10}$ |
| $[0+2.7726+2(0.2625+\text{their }0.6595+1.2032+1.9044)]$ | M1 | For structure of trapezium rule $[\ldots]$; no errors in ordinates allowed |
| $=\frac{1}{10}(10.8318)=1.083$ (3 dp) | A1 | Anything that rounds to $1.083$ |

## Part (c) — Way 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=\ln x \Rightarrow \frac{du}{dx}=\frac{1}{x}$; $\frac{dv}{dx}=x^2 \Rightarrow v=\frac{1}{3}x^3$ | — | — |
| $x^2\ln x \to \pm\lambda x^3\ln x - \int \mu x^3\left(\frac{1}{x}\right)\{dx\}$ or $\pm\lambda x^3\ln x - \int \mu x^2\{dx\}$ | M1 | $\lambda,\mu>0$ |
| $=\frac{x^3}{3}\ln x - \int\frac{x^3}{3}\left(\frac{1}{x}\right)\{dx\}$ | A1 | $\frac{x^3}{3}\ln x - \int\frac{x^3}{3}\cdot\frac{1}{x}\{dx\}$, simplified or un-simplified |
| $=\frac{x^3}{3}\ln x - \frac{x^3}{9}$ | A1 | Simplified or un-simplified |
| $\text{Area}=\left[\frac{x^3}{3}\ln x-\frac{x^3}{9}\right]_1^2=\left(\frac{8}{3}\ln 2-\frac{8}{9}\right)-\left(0-\frac{1}{9}\right)$ | dM1 | Dependent on previous M; applies limits 2 and 1, subtracts correct way round |
| $=\frac{8}{3}\ln 2-\frac{7}{9}$ | A1 oe cso | Or $\frac{1}{9}(24\ln 2-7)$ |

## Part (c) — Way 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=x^2$, $\frac{dv}{dx}=\ln x \Rightarrow v=x\ln x - x$ | — | — |
| $I=x^2(x\ln x-x)-\int 2x(x\ln x - x)dx$ leading to $3I=x^2(x\ln x-x)+\int 2x^2\{dx\}$ | M1 | Full method of applying $u=x^2$, $v'=\ln x$ giving $\pm\lambda x^2(x\ln x-x)\pm\mu\int x^2\{dx\}$ |
| $I=\frac{1}{3}x^2(x\ln x-x)+\frac{1}{3}\int 2x^2\{dx\}$ | A1 | Simplified or un-simplified |
| $=\frac{x^3}{3}\ln x-\frac{x^3}{9}$ | A1 | Simplified or un-simplified |
| Apply limits as above | M1 A1 | Then award dM1A1 in the same way as above |

## Question 2 Notes

| Note | Guidance |
|---|---|
| (b) B1 | Outside brackets $\frac{1}{2}\times(0.2)$ or $\frac{1}{2}\times\frac{1}{5}$ or $\frac{1}{10}$ or equivalent |
| (b) Note | No errors allowed (omission of a $y$-ordinate or extra/repeated $y$-ordinate) |
| (b) Note | Full marks can be gained using an incorrect part (a) answer of $0.6594$ |
| (c) A1 | Exact answer must be two-term expression in form $a\ln b+c$ |
| (c) Note | Give final A0 for $\frac{8\ln 2-\ln 1}{3}-\frac{7}{9}$ or $\frac{8\ln 2}{3}-\frac{1}{3}\ln 1-\frac{7}{9}$ or $\frac{8\ln 2}{3}-\frac{8}{9}+\frac{1}{9}$ |
| (c) Note | $\left[\frac{x^3}{3}\ln x-\frac{x^3}{9}\right]_1^2$ followed by awrt $1.07$ with no correct answer seen is dM1A0 |
| (c) Note | Give dM0A0 for adding rather than subtracting limits |
| (c) Note | Allow dM1A0 for $\left(\frac{8}{3}\ln 2-\frac{8}{9}\right)-\left(0+\frac{1}{9}\right)$ |
| (c) SC | Candidate using $u=\ln x$, $\frac{dv}{dx}=x^2$ but making only one error when applying by parts: award Special Case 1st M1 |