| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Show that integral equals expression |
| Difficulty | Moderate -0.3 This is a straightforward single application of integration by parts with a standard function pair (x and ln x), followed by routine evaluation of definite integral limits. It requires less problem-solving than a typical A-level question since the method is directly indicated and involves only one integration by parts step with no algebraic complications. |
| Spec | 1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = \ln x, u' = \frac{1}{x}, v' = x, v = \frac{1}{2}x^2\) | M1 | |
| \(I = [\frac{1}{2}x^2 \ln x]_1^2 - \int_1^2 \frac{1}{2}x \, dx\) | A1 | |
| \(= [\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2]_1^2\) | M1 A1 | |
| \(= (2 \ln 2 - 1) - (0 - \frac{1}{4}) = 2 \ln 2 - \frac{3}{4}\) | M1 A1 | (6 marks) |
$u = \ln x, u' = \frac{1}{x}, v' = x, v = \frac{1}{2}x^2$ | M1 |
$I = [\frac{1}{2}x^2 \ln x]_1^2 - \int_1^2 \frac{1}{2}x \, dx$ | A1 |
$= [\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2]_1^2$ | M1 A1 |
$= (2 \ln 2 - 1) - (0 - \frac{1}{4}) = 2 \ln 2 - \frac{3}{4}$ | M1 A1 | (6 marks)\begin{enumerate}
\item Use integration by parts to show that
\end{enumerate}
$$\int _ { 1 } ^ { 2 } x \ln x \mathrm {~d} x = 2 \ln 2 - \frac { 3 } { 4 }$$
\hfill \mbox{\textit{Edexcel C4 Q1 [6]}}