| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Square root substitution: definite integral |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear guidance at each step. Part (i) is routine trapezium rule application, part (ii) is a standard substitution with the substitution explicitly given (not requiring students to identify it), and part (iii) involves simple algebraic manipulation to match a given form. While it requires competent technique across multiple areas, there's no novel problem-solving or insight required—students follow the prescribed method at each stage. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 2\) | B1 1.1 | — |
| \(\dfrac{h}{2}\left[\dfrac{1}{2} + \dfrac{1}{4} + 2\left(\dfrac{1}{2+\sqrt{2}}\right)\right]\) | M1 2.1 | Use of correct formula with correct (exact) \(y\)-values with their \(h\); condone one error in values |
| \(I \approx \dfrac{3}{4} + \dfrac{2}{2+\sqrt{2}}\) | A1 1.1 | — |
| \(\dfrac{1}{2+\sqrt{2}} = \dfrac{(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} = \dfrac{2-\sqrt{2}}{2}\) | M1 3.1a | Correct method for rationalising denominator of their surd with correct simplification |
| \(I \approx \dfrac{3}{4} + (2 - \sqrt{2}) = \dfrac{11}{4} - \sqrt{2}\) | A1 2.2a [5] | AG — at least one step of intermediate working (from application of trapezium rule to given result); must be convincing as AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = u^2 \Rightarrow dx = 2u\, du\) | M1* 3.1a | An attempt at integration by substitution — allow any genuine attempt; limits not required for first four marks |
| \(\displaystyle\int_0^4 \frac{dx}{2+\sqrt{x}} = \int_0^2 \frac{2u}{2+u}\, du\) | A1 1.1 | Correct integral in terms of \(u\) |
| \(= 2\displaystyle\int_0^2 \frac{2+u-2}{2+u}\, du = 2\int_0^2 1 - \dfrac{2}{2+u}\, du\) | Dep*M1 2.1 | Re-writes integral in form \(\displaystyle\int a + \frac{b}{1+u}\, du\); or use \(t = 2+u\) to obtain \(\displaystyle\int a + \frac{b}{t}\, dt\) |
| \(= 2\left[u - 2\ln(2+u)\right]_0^2\) | A1ft 1.1 | Correctly integrates \(\displaystyle\int a + \frac{b}{1+u}\, du\); \(\displaystyle\int 2 - \frac{4}{t}\, dt = 2t - 4\ln t\) |
| \(= 2\{(2 - 2\ln(2+2)) - (0 - 2\ln(2+0))\}\) | M1 1.1 | Uses correct limits correctly (dependent on both previous M marks) |
| \(= 2(2 - 2\ln 2)\) | A1 2.2a [6] | oe e.g. \(4 - 4\ln 4 + 4\ln 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{11}{4} - \sqrt{2} \approx 2(2 - 2\ln 2)\) | M1 1.1a | Setting given result approx. equal to their (ii) |
| \(\ln 2 \approx \dfrac{5}{16} + \dfrac{\sqrt{2}}{4}\) | A1 2.2a [2] | \(k = \dfrac{5}{16}\) |
## Question 5(i):
$h = 2$ | B1 1.1 | —
$\dfrac{h}{2}\left[\dfrac{1}{2} + \dfrac{1}{4} + 2\left(\dfrac{1}{2+\sqrt{2}}\right)\right]$ | M1 2.1 | Use of correct formula with correct (exact) $y$-values with their $h$; condone one error in values
$I \approx \dfrac{3}{4} + \dfrac{2}{2+\sqrt{2}}$ | A1 1.1 | —
$\dfrac{1}{2+\sqrt{2}} = \dfrac{(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} = \dfrac{2-\sqrt{2}}{2}$ | M1 3.1a | Correct method for rationalising denominator of their surd with correct simplification
$I \approx \dfrac{3}{4} + (2 - \sqrt{2}) = \dfrac{11}{4} - \sqrt{2}$ | A1 2.2a [5] | **AG** — at least one step of intermediate working (from application of trapezium rule to given result); must be convincing as AG
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## Question 5(ii):
$x = u^2 \Rightarrow dx = 2u\, du$ | M1* 3.1a | An attempt at integration by substitution — allow any genuine attempt; limits not required for first four marks
$\displaystyle\int_0^4 \frac{dx}{2+\sqrt{x}} = \int_0^2 \frac{2u}{2+u}\, du$ | A1 1.1 | Correct integral in terms of $u$
$= 2\displaystyle\int_0^2 \frac{2+u-2}{2+u}\, du = 2\int_0^2 1 - \dfrac{2}{2+u}\, du$ | Dep*M1 2.1 | Re-writes integral in form $\displaystyle\int a + \frac{b}{1+u}\, du$; or use $t = 2+u$ to obtain $\displaystyle\int a + \frac{b}{t}\, dt$
$= 2\left[u - 2\ln(2+u)\right]_0^2$ | A1ft 1.1 | Correctly integrates $\displaystyle\int a + \frac{b}{1+u}\, du$; $\displaystyle\int 2 - \frac{4}{t}\, dt = 2t - 4\ln t$
$= 2\{(2 - 2\ln(2+2)) - (0 - 2\ln(2+0))\}$ | M1 1.1 | Uses correct limits correctly (dependent on both previous M marks)
$= 2(2 - 2\ln 2)$ | A1 2.2a [6] | oe e.g. $4 - 4\ln 4 + 4\ln 2$
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## Question 5(iii):
$\dfrac{11}{4} - \sqrt{2} \approx 2(2 - 2\ln 2)$ | M1 1.1a | Setting given result approx. equal to their (ii)
$\ln 2 \approx \dfrac{5}{16} + \dfrac{\sqrt{2}}{4}$ | A1 2.2a [2] | $k = \dfrac{5}{16}$
---5 (i) Use the trapezium rule, with two strips of equal width, to show that
$$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x \approx \frac { 11 } { 4 } - \sqrt { 2 }$$
(ii) Use the substitution $x = u ^ { 2 }$ to find the exact value of
$$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { x } } \mathrm {~d} x$$
(iii) Using your answers to parts (i) and (ii), show that
$$\ln 2 \approx k + \frac { \sqrt { 2 } } { 4 }$$
where $k$ is a rational number to be determined.
\hfill \mbox{\textit{OCR H240/03 2018 Q5 [13]}}