| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Improper integral evaluation |
| Difficulty | Standard +0.3 Part (i) is a routine polynomial integration well below average difficulty. Part (ii) introduces an improper integral with an infinite limit, requiring understanding of limits, but the integrand 1/x^4 is straightforward to integrate and converges simply. This is slightly above average for C2 due to the improper integral concept, but the execution is mechanical. |
| Spec | 1.08d Evaluate definite integrals: between limits4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks |
|---|---|
| \(= [\frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x]_1^4\) | M1 A2 |
| \(= (\frac{64}{3} - 40 + 16) - (\frac{1}{3} - \frac{5}{2} + 4) = -\frac{9}{2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \lim_{k\to\infty} [-\frac{1}{3}x^{-3}]_k^{-1}\) | M2 A1 | |
| \(= \lim_{k\to\infty} \{\frac{1}{3} - (-\frac{1}{3k^3})\}\) | M1 | |
| \(= \lim_{k\to\infty} (\frac{1}{3} + \frac{1}{3k^3}) = \frac{1}{3}\) | A1 | (10) |
# Question 6:
## Part (i):
$= [\frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x]_1^4$ | M1 A2 |
$= (\frac{64}{3} - 40 + 16) - (\frac{1}{3} - \frac{5}{2} + 4) = -\frac{9}{2}$ | M1 A1 |
## Part (ii):
$= \lim_{k\to\infty} [-\frac{1}{3}x^{-3}]_k^{-1}$ | M2 A1 |
$= \lim_{k\to\infty} \{\frac{1}{3} - (-\frac{1}{3k^3})\}$ | M1 |
$= \lim_{k\to\infty} (\frac{1}{3} + \frac{1}{3k^3}) = \frac{1}{3}$ | A1 | **(10)**
---6. Evaluate\\
(i) $\quad \int _ { 1 } ^ { 4 } \left( x ^ { 2 } - 5 x + 4 \right) \mathrm { d } x$,\\
(ii) $\int _ { - \infty } ^ { - 1 } \frac { 1 } { x ^ { 4 } } \mathrm {~d} x$.\\
\hfill \mbox{\textit{OCR C2 Q6 [10]}}