| 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 Part (i) is a straightforward application of the reverse chain rule for a linear substitution, requiring only pattern recognition. Part (ii) explicitly provides the substitution and involves standard technique with limits adjustment. Both are routine C3-level exercises with no problem-solving insight required, making this slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(u = 2x-3\), \(\frac{du}{dx} = 2 \Rightarrow dx = \frac{1}{2}du\) | M1 A1 | or B3 cao |
| \(= \int \frac{1}{2}u^7 \, du = \frac{u^8}{2 \times 8} = \frac{1}{16}(2x-3)^8 + c\) | A1 | |
| 3 | ||
| (ii) The substitution \(u = x^2+1\) gives \(\frac{du}{dx} = 2x\) | M1 | Using sub |
| \(\Rightarrow \int_1^2 x(x^2+1)^3 \, dx = \int_2^5 \frac{1}{2}u^3 \, du\) | A1 | Correct int |
| \(= \left[\frac{u^4}{8}\right]_2^5\) | A1 | Correct limits |
| \(= \frac{609}{8} (= 76\frac{1}{8})\) | A1 A1 | Int, Ans |
| 5 |
**(i)** $\int(2x-3)^7 \, dx$
Let $u = 2x-3$, $\frac{du}{dx} = 2 \Rightarrow dx = \frac{1}{2}du$ | M1 A1 | or B3 cao
$= \int \frac{1}{2}u^7 \, du = \frac{u^8}{2 \times 8} = \frac{1}{16}(2x-3)^8 + c$ | A1 |
| 3 |
**(ii)** The substitution $u = x^2+1$ gives $\frac{du}{dx} = 2x$ | M1 | Using sub
$\Rightarrow \int_1^2 x(x^2+1)^3 \, dx = \int_2^5 \frac{1}{2}u^3 \, du$ | A1 | Correct int
$= \left[\frac{u^4}{8}\right]_2^5$ | A1 | Correct limits
$= \frac{609}{8} (= 76\frac{1}{8})$ | A1 A1 | Int, Ans
| 5 |6 (i) Find $\int ( 2 x - 3 ) ^ { 7 } \mathrm {~d} x$.\\
(ii) Use the substitution $u = x ^ { 2 } + 1$, or otherwise, to find $\int _ { 1 } ^ { 2 } x \left( x ^ { 2 } + 1 \right) ^ { 3 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR MEI C3 Q6 [8]}}