| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Show that integral equals expression |
| Difficulty | Standard +0.3 This is a straightforward integration by parts question with a definite integral. Students need to apply the standard IBP formula once (choosing u = ln x, dv = x³ dx), evaluate the resulting integral (x⁴/4), and substitute the limits e² and 1. The algebra is clean with nice cancellation when ln(1) = 0. While it requires careful execution and the 'show all working' instruction adds slight pressure, it's a standard textbook exercise with no conceptual surprises—slightly easier than average for A-level. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int x^3 \ln x\, dx = \frac{x^4}{4}\ln x - \int \frac{x^4}{4} \times \frac{1}{x}\, dx\) | M1 | 1.1b — integrates by parts the right way round; look for \(kx^4 \ln x - \int kx^4 \times \frac{1}{x}\, dx\), \(k>0\) |
| \(= \frac{x^4}{4}\ln x - \frac{x^4}{16}\ (+c)\) | M1, A1 | 1.1b, 1.1b — correct method to integrate \(\int kx^4 \times \frac{1}{x}\, dx \to cx^4\) |
| \(\int_1^{e^2} x^3 \ln x\, dx = \left[\frac{x^4}{4}\ln x - \frac{x^4}{16}\right]_1^{e^2} = \left(\frac{e^8}{4}\ln e^2 - \frac{e^8}{16}\right) - \left(-\frac{1^4}{16}\right)\) | M1 | 2.1 — substitutes 1 and \(e^2\), uses \(\ln e^2 = 2\) |
| \(= \frac{7}{16}e^8 + \frac{1}{16}\) | A1 | 1.1b — allow \(0.4375e^8 + 0.0625\); NOT ISW: \(7e^8 + 1\) is A0 |
# Question 12:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int x^3 \ln x\, dx = \frac{x^4}{4}\ln x - \int \frac{x^4}{4} \times \frac{1}{x}\, dx$ | M1 | 1.1b — integrates by parts the right way round; look for $kx^4 \ln x - \int kx^4 \times \frac{1}{x}\, dx$, $k>0$ |
| $= \frac{x^4}{4}\ln x - \frac{x^4}{16}\ (+c)$ | M1, A1 | 1.1b, 1.1b — correct method to integrate $\int kx^4 \times \frac{1}{x}\, dx \to cx^4$ |
| $\int_1^{e^2} x^3 \ln x\, dx = \left[\frac{x^4}{4}\ln x - \frac{x^4}{16}\right]_1^{e^2} = \left(\frac{e^8}{4}\ln e^2 - \frac{e^8}{16}\right) - \left(-\frac{1^4}{16}\right)$ | M1 | 2.1 — substitutes 1 and $e^2$, uses $\ln e^2 = 2$ |
| $= \frac{7}{16}e^8 + \frac{1}{16}$ | A1 | 1.1b — allow $0.4375e^8 + 0.0625$; NOT ISW: $7e^8 + 1$ is A0 |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
Show that
$$\int _ { 1 } ^ { \mathrm { e } ^ { 2 } } x ^ { 3 } \ln x \mathrm {~d} x = a \mathrm { e } ^ { 8 } + b$$
where $a$ and $b$ are rational constants to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q12 [5]}}