| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/(a²+x²) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard inverse trig integration formulas. Part (i) requires completing the square to get the arctan form, and part (ii) is direct application of arcsin integration with a mean value calculation. Both are textbook exercises with clear methods and minimal problem-solving required, though slightly above average difficulty due to being Further Maths content. |
| Spec | 4.08f Integrate using partial fractions4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 - 2x + 10 = (x-1)^2 + 9\) | M1 | Completing the square |
| \(\Rightarrow \int_1^4 \frac{1}{(x-1)^2+9}dx = \frac{1}{3}[\tan^{-1}(\frac{x-1}{3})]_1^4\) | A1 | Ignore limits |
| \(= \frac{1}{3}(\frac{\pi}{4} - 0) = \frac{\pi}{12}\) | A1 [3] | |
| (ii) Mean value \(= \frac{1}{(\frac{1}{2}-0)} \int_0^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx\) | M1 | Uses formula for mean value |
| \(= 2[\sin^{-1}x]_0^{\frac{1}{2}} = 2(\frac{\pi}{6} - 0)\) | M1 [3] | Uses formula for integral and attempts substitution of limits |
| \(= \frac{1}{3}\pi\) | A1 [3] |
**(i)** $x^2 - 2x + 10 = (x-1)^2 + 9$ | M1 | Completing the square
$\Rightarrow \int_1^4 \frac{1}{(x-1)^2+9}dx = \frac{1}{3}[\tan^{-1}(\frac{x-1}{3})]_1^4$ | A1 | Ignore limits
$= \frac{1}{3}(\frac{\pi}{4} - 0) = \frac{\pi}{12}$ | A1 [3] |
**(ii)** Mean value $= \frac{1}{(\frac{1}{2}-0)} \int_0^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx$ | M1 | Uses formula for mean value
$= 2[\sin^{-1}x]_0^{\frac{1}{2}} = 2(\frac{\pi}{6} - 0)$ | M1 [3] | Uses formula for integral and attempts substitution of limits
$= \frac{1}{3}\pi$ | A1 [3]
---4 In this question you must show detailed reasoning.\\
Find the exact value of each of the following.\\
(i) $\int _ { 1 } ^ { 4 } \frac { 1 } { x ^ { 2 } - 2 x + 10 } \mathrm {~d} x$\\
(ii) The mean value of $\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$ in the interval $[ 0,0.5 ]$
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q4 [6]}}