| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with simple linear/polynomial substitution |
| Difficulty | Moderate -0.3 Both parts are standard C3 integration exercises. Part (i) is a direct recognition of derivative-over-function form leading to ln|x²+1|, while part (ii) requires simple algebraic manipulation (polynomial division or rewriting as 2(x+1)-2) before integration. These are textbook-level questions testing routine substitution/recognition skills with straightforward exact answers, making them slightly easier than average A-level questions. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(= \left[\ln(x^2+1)\right]_0^1\) | M2, A1 | \([\ln(x^2+1)]\); cao (must be exact) |
| \(= \ln 2\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \frac{2x}{x^2+1}\,dx = \int_1^2 \frac{1}{u}\,du\) | M1 | \(\int \frac{1}{u}\,du\) or \(\left[\ln(1+x^2)\right]_0^1\) with correct limits |
| \(= [\ln u]_1^2\) | A1 | |
| \(= \ln 2\) | A1 [3] | cao (must be exact) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \frac{2x+2-2}{x+1}\,dx = \int_0^1\left(2 - \frac{2}{x+1}\right)dx\) | 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/Working | Marks | Guidance |
| \(= \int_1^2 \frac{2(u-1)}{u}\,du\) | M1, B1 | Substituting \(u=x+1\) and \(du=dx\) (or \(du/dx=1\)) *and* correct limits; \(2(u-1)/u\) |
| \(= \int_1^2\left(2 - \frac{2}{u}\right)du\) | 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): $\int_0^1 \frac{2x}{x^2+1}\,dx$
| Answer/Working | Marks | Guidance |
|---|---|---|
| $= \left[\ln(x^2+1)\right]_0^1$ | M2, A1 | $[\ln(x^2+1)]$; cao (must be exact) |
| $= \ln 2$ | [3] | |
**Or** let $u = x^2+1$, $du = 2x\,dx$:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \frac{2x}{x^2+1}\,dx = \int_1^2 \frac{1}{u}\,du$ | M1 | $\int \frac{1}{u}\,du$ or $\left[\ln(1+x^2)\right]_0^1$ with correct limits |
| $= [\ln u]_1^2$ | A1 | |
| $= \ln 2$ | A1 [3] | cao (must be exact) |
### Part (ii): $\int_0^1 \frac{2x}{x+1}\,dx$
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \frac{2x+2-2}{x+1}\,dx = \int_0^1\left(2 - \frac{2}{x+1}\right)dx$ | 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** let $u = x+1$, $du = dx$:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $= \int_1^2 \frac{2(u-1)}{u}\,du$ | M1, B1 | Substituting $u=x+1$ and $du=dx$ (or $du/dx=1$) *and* correct limits; $2(u-1)/u$ |
| $= \int_1^2\left(2 - \frac{2}{u}\right)du$ | 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.
\begin{displayquote}
(i) $\int _ { 0 } ^ { 1 } \frac { 2 x } { x ^ { 2 } + 1 } \mathrm {~d} x$\\
(ii) $\int _ { 0 } ^ { 1 } \frac { 2 x } { x + 1 } \mathrm {~d} x$
\end{displayquote}
\hfill \mbox{\textit{OCR MEI C3 Q4 [8]}}