| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with simple linear/polynomial substitution |
| Difficulty | Moderate -0.8 Both parts are standard integration exercises requiring straightforward substitutions (u = x² + 1 for part i, u = x + 1 for part ii) that students would have practiced extensively. Part (i) is direct recognition of f'(x)/f(x) form, while part (ii) requires simple polynomial division before integrating. These are textbook-level exercises with clear methods and no problem-solving insight required, making them easier than average A-level questions. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \frac{2x}{x^2+1}dx = \left[\ln(x^2+1)\right]_0^1\) | M2, A1 | \([\ln(x^2+1)]\); cao (must be exact) |
| \(= \ln 2\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(u = x^2+1\), \(du = 2x\,dx \Rightarrow \int_0^1 \frac{2x}{x^2+1}dx = \int_1^2 \frac{1}{u}du\) | M1 | \(\int \frac{1}{u}du\) |
| \(= \left[\ln u\right]_1^2\) | A1 | or \(\left[\ln(1+x^2)\right]_0^1\) with correct limits |
| \(= \ln 2\) | A1 [3] | cao (must be exact) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \frac{2x}{x+1}dx = \int_0^1 \frac{2x+2-2}{x+1}dx = \int_0^1\left(2-\frac{2}{x+1}\right)dx\) | M1, A1, A1 | dividing by \((x+1)\); \(2, -2/(x+1)\) |
| \(= \left[2x - 2\ln(x+1)\right]_0^1\) | A1 | |
| \(= 2 - 2\ln 2\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(u = x+1\), \(\Rightarrow du = dx\) | M1 | substituting \(u=x+1\) and \(du=dx\) |
| \(= \int_1^2 \frac{2(u-1)}{u}du\) | B1 | correct limits used for \(u\) or \(x\) |
| \(= \int_1^2\left(2-\frac{2}{u}\right)du\) | M1 | dividing through by \(u\) |
| \(= \left[2u - 2\ln u\right]_1^2\) | A1 | \(2u-2\ln u\); allow ft on \((u-1)/u\) (i.e. with 2 omitted) |
| \(= 4 - 2\ln 2 - (2-2\ln 1) = 2-2\ln 2\) | A1 [5] | o.e. cao (must be exact) |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{2x}{x^2+1}dx = \left[\ln(x^2+1)\right]_0^1$ | M2, A1 | $[\ln(x^2+1)]$; cao (must be exact) |
| $= \ln 2$ | [3] | |
**Or (substitution method):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $u = x^2+1$, $du = 2x\,dx \Rightarrow \int_0^1 \frac{2x}{x^2+1}dx = \int_1^2 \frac{1}{u}du$ | M1 | $\int \frac{1}{u}du$ |
| $= \left[\ln u\right]_1^2$ | A1 | or $\left[\ln(1+x^2)\right]_0^1$ with correct limits |
| $= \ln 2$ | A1 [3] | cao (must be exact) |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{2x}{x+1}dx = \int_0^1 \frac{2x+2-2}{x+1}dx = \int_0^1\left(2-\frac{2}{x+1}\right)dx$ | M1, A1, A1 | dividing by $(x+1)$; $2, -2/(x+1)$ |
| $= \left[2x - 2\ln(x+1)\right]_0^1$ | A1 | |
| $= 2 - 2\ln 2$ | A1 [5] | |
**Or (substitution):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $u = x+1$, $\Rightarrow du = dx$ | M1 | substituting $u=x+1$ and $du=dx$ |
| $= \int_1^2 \frac{2(u-1)}{u}du$ | B1 | correct limits used for $u$ or $x$ |
| $= \int_1^2\left(2-\frac{2}{u}\right)du$ | M1 | dividing through by $u$ |
| $= \left[2u - 2\ln u\right]_1^2$ | A1 | $2u-2\ln u$; allow ft on $(u-1)/u$ (i.e. with 2 omitted) |
| $= 4 - 2\ln 2 - (2-2\ln 1) = 2-2\ln 2$ | A1 [5] | o.e. cao (must be exact) |
---4 Evaluate the following integrals, giving your answers in exact form.\\
(i) $\int _ { 0 } ^ { 1 } \frac { 2 x } { x ^ { 2 } + 1 } \mathrm {~d} x$.\\
(ii) $\int _ { 0 } ^ { 1 } \frac { 2 x } { x + 1 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR MEI C3 2010 Q4 [8]}}