A2 June 2024 Paper 1 Q7
7
Use an appropriate limit argument to evaluate this integral. [4]
| Scheme | Marks | AO |
|---|---|---|
| Because \(\dfrac{1}{\sqrt[3]{x - 2}}\) is not defined when \(x = 2\) | B1 | 2.4 |
| [1] |
Notes
B1: isw. Must explicitly refer to \(x = 2\). See appendix.
Appendix: exemplar responses for Q7a
| Response | Mark |
|---|---|
| \(\dfrac{1}{\sqrt[3]{x - 2}}\) is undefined when \(x = 2\) | B1 |
| There is an asymptote at \(x = 2\) | B1 |
| The function is not continuous at \(x = 2\) | B1 |
| When \(x = 2\) the denominator will be 0 and \(\frac{1}{0}\) is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computed | B1 |
| When \(x = 2\) we get 1 divided by 0 which is undefined/impossible/invalid/improper/indeterminate/an error/cannot be computed | B1 |
| When \(x = 2\) we get \(\frac{1}{0}\) which makes the integral undefined/impossible/invalid/improper/indeterminate/an error/incomputable | B1 |
| When \(x = 2\) the denominator of the function is 0 and you cannot divide by 0 | B1 |
| When \(x = 2\) it becomes \(\frac{1}{0}\) = undefined | B1 |
| When \(x = 2\) we get \(\frac{1}{0}\) which tends to infinity | B0 |
| When \(x = 2\) we get \(\frac{1}{0}\) which is divergent | B0 |
| When \(x = 2\) we get \(\frac{1}{0}\) which has no solutions | B0 |
| When \(x = 2\) we get 1 divided by 0 which is an error function | B0 |
| When \(x \lt 2\) the value is complex/not a real number | B0 |
| It is undefined when \(x \to 2\) | B0 |
| It is undefined when \(x = 2\) | B1 |
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int \frac{1}{(x - 2)^{\frac{1}{3}}}\,\mathrm{d}x = \frac{3}{2}(x - 2)^{\frac{2}{3}}\) | B1* | 2.1 |
| \(\displaystyle\lim_{a \to 2}\int_1^a \frac{1}{\sqrt[3]{x - 2}}\,\mathrm{d}x\) or \(\displaystyle\lim_{a \to 2}\left[\frac{3}{2}(a - 2)^{\frac{2}{3}}\right]_1^a\) | B1 | 2.1 |
| \(\displaystyle\lim_{a \to 2}\left[\frac{3}{2}(a - 2)^{\frac{2}{3}}\right] = 0\) \(\displaystyle\left[\int_1^2 \frac{1}{\sqrt[3]{x - 2}}\,\mathrm{d}x =\right] 0 - \frac{3}{2}(1 - 2)^{\frac{2}{3}}\) | B1 | 2.4 |
| \(= -\dfrac{3}{2}\) | B1dep | 2.2a |
| [4] |
Notes
B1: Introducing an algebraic limit for ‘2’ in original integral or in integral of form \(k(a - 2)^{2/3}\). Soi by next B1. Do not allow \(x\) as a limit.
B1: Clear limit argument used for \(k(a - 2)^{2/3}\) as \(a \to 2\). Do not allow \(x\) used for \(a\). = or \(\to\) must be used correctly.
B1dep: Integral must have been explicitly evaluated for both limits. Do not accept “\(\to -\frac{3}{2}\)”. Withhold if two limit arguments considered.
Alternative method
| Scheme | Marks |
|---|---|
| Let \(u = x - 2\) \(\displaystyle\int \frac{1}{\sqrt[3]{u}}\,\mathrm{d}u = \frac{3}{2}u^{\frac{2}{3}}\) | B1* |
| \(\displaystyle\lim_{a \to 0}\int_{-1}^a \frac{1}{\sqrt[3]{u}}\,\mathrm{d}u\) or \(\displaystyle\lim_{a \to 0}\left[\frac{3}{2}u^{\frac{2}{3}}\right]_{-1}^a\) | B1 |
| \(\displaystyle\lim_{a \to 0}\left[\frac{3}{2}a^{\frac{2}{3}}\right] = 0\) \(\displaystyle\left[\int_{-1}^0 \frac{1}{\sqrt[3]{u}}\,\mathrm{d}u =\right] 0 - \frac{3}{2}(-1)^{\frac{2}{3}}\) | B1 |
| \(= -\dfrac{3}{2}\) | B1dep |
| [4] |
B1*: Correct substitution and integration
B1: Introducing an algebraic limit for ‘0’ in original integral in terms of \(u\) or in integral of form \(ku^{2/3}\). Soi by next B1. Do not allow \(u\) as a limit.
B1: Clear limit argument used for \(k(a)^{2/3}\) as \(a \to 0\). Do not allow \(x\) used for \(a\). = or \(\to\) must be used correctly.
B1dep: Integral must have been explicitly evaluated for both limits. Do not accept “\(\to -\frac{3}{2}\)”. Withhold if two limit arguments considered.