# Question 13:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| awrt $0.3799$ | B1 | Should be in the table or given as answer – not just appear in trapezium rule |

**(1 mark)**

---

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Strip width of $\frac{e^2-e}{2}$ or correct equivalent e.g. $\frac{e^2+e}{2}-e$, or $e^2-\frac{e^2+e}{2}$ | B1 | Also accept awrt 2.34 o.e. May be stated as $h=$ the values above, or used correctly so $\frac{e^2-e}{4}$ etc or 1.17 may be seen |
| Area $= \frac{1}{2}\times\left(\frac{e^2-e}{2}\right)(1+2\times 0.3799+0)$ | M1 | For **correct application** of trapezium rule – requires correct ft bracket $\ldots\times(1+2\times$ their answer to $(a)+0)$ |
| $=$ awrt $2.055$ | A1 | |

**(3 marks)**

---

## Part (c) Way 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int(\ln x)^2\,dx = \int 1\times(\ln x)^2\,dx = x(\ln x)^2 - \int x\times\frac{2\ln x}{x}\,dx$ | M1 | One correct application of integration by parts the correct way around. Accept $\int(\ln x)^2\,dx = x(\ln x)^2 - \int x\times\frac{A\ln x}{x}\,dx$ |
| $= x(\ln x)^2 - \int 2\ln x\,dx$ | A1 | |
| $= x(\ln x)^2 - 2x\ln x + \int 2\,dx$ | dM1 | A second application of integration by parts the correct way around. Accept $= x(\ln x)^2 - Ax\ln x \pm \int B\,dx$. Accept the answer to $\int\ln x\,dx$ just written down as $x\ln x - x$ |
| $= x(\ln x)^2 - 2x\ln x + 2x\,(+c)$ | A1* | Correct solution only, with or without $c$ |

**(4 marks)**

---

## Part (c) Way 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\int(\ln x)^2\,dx = \int(\ln x)\times(\ln x)\,dx = \ln x(x\ln x - x) - \int\frac{1}{x}\times(x\ln x - x)\,dx$ | M1 | One correct application of integration by parts the correct way around |
| $= \ln x(x\ln x-x) - \int(\ln x - 1)\,dx$ | A1 | |
| $= \ln x(x\ln x-x)-(x\ln x - x - x)$ | dM1 | A second application of integration by parts the correct way around. Accept the answer to $\int\ln x\,dx$ just written down as $x\ln x - x$ and substituted into their expression |
| $= x(\ln x)^2 - 2x\ln x + 2x\,(+c)$ | A1* | Correct solution only, with or without $c$ |

**(4 marks)**

---

## Part (c) Way 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Use $u=\ln x$ substitution: $\int u^2 e^u\,du = u^2e^u - \int 2ue^u\,du$ | M1 | Uses substitution and performs one correct application of integration by parts the correct way around |
| $= u^2e^u - \left[2ue^u - \int 2e^u\,du\right]$ | A1 | After second integration by parts |
| $= u^2e^u - 2ue^u + 2e^u + k$ | dM1 | Final integration and returns to $x$ |
| $= x(\ln x)^2 - 2x\ln x + 2x\,(+c)$ | A1* | Correct solution only, with or without $c$ |

**(4 marks)**

---

## Part (d):

| Answer | Mark | Guidance |
|--------|------|----------|
| Volume $= \int\pi y^2\,dx = \int\pi(2-\ln x)^2\,dx$ | B1 | Needs $\pi$ and integral symbol but not limits; can condone missing $dx$ |
| $\int(2-\ln x)^2\,dx = \int 4 - 4\ln x + (\ln x)^2\,dx$ | M1 | Multiplies out $(2-\ln x)^2$ to $\lambda + \mu\ln x + \nu(\ln x)^2$ where $\lambda$, $\mu$, $\nu$ are non-zero constants |
| Correct integration of at least two of their three terms | M1 | Needs attempt to multiply out to at least $\lambda + \nu(\ln x)^2$ and attempt to integrate. Look for $\lambda x \pm\mu(x\ln x - x)\pm\nu\!\left(x(\ln x)^2-2x\ln x+2x\right)$ with **two of the three** terms integrated correctly (if $\mu=0$ could score M0M1) |
| $= 4x - 4(x\ln x - x) + x(\ln x)^2 - 2x\ln x + 2x\,(+c)$ | A1 | Correct answer, accept unsimplified and isw |
| Volume $= \pi\!\left[4x - 4(x\ln x-x)+x(\ln x)^2 - 2x\ln x+2x\right]_e^{e^2}$ | ddM1 | Attempts to substitute both correct limits into the result of their integral. Both previous M marks must have been scored. Allow M1 if small slips simplifying before use of limits |
| $= 2\pi e^2 - 5\pi e = \pi e(2e-5)$ | A1 | Correct solution only |

**(6 marks)**

**Special case:** For those who misread and think $y=(\ln x)^2$ so Volume $=\int\pi(\ln x)^4\,dx$ — first B1 may be exceptionally awarded. This gains 1/6 marks.

---

**(14 marks total)**