June 2025 Paper 1 Q13
13.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Show that\[\int_0^2 \frac{x}{(2x+1)^3}\,\mathrm{d}x = \frac{2}{25}\](5)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,\mathrm{d}x = \dfrac{x(2x+1)^{-2}}{-4} + \displaystyle\int \dfrac{(2x+1)^{-2}}{4}\,(\mathrm{d}x)\) | M1 | 3.1a |
| \(= \ldots + \dfrac{(2x+1)^{-1}}{-8}\) | dM1 | 1.1b |
| \(-\dfrac{x}{4(2x+1)^2} - \dfrac{1}{8(2x+1)}\) | A1 | 1.1b |
| e.g. \(= \left(-\dfrac{2}{4(2 \times 2 + 1)^2} - \dfrac{1}{8(2 \times 2 + 1)}\right) - \left(-\dfrac{0}{4(2 \times 0 + 1)^2} - \dfrac{1}{8(2 \times 0 + 1)}\right)\) | ddM1 | 1.1b |
| e.g. \(-\dfrac{2}{100} - \dfrac{1}{40} + \dfrac{1}{8} = \dfrac{2}{25}\) * | A1* | 2.1 |
| (5) | ||
| (5 marks) |
Notes
13 Alt I
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,\mathrm{d}x = \displaystyle\int \dfrac{u - 1}{4u^3}\,(\mathrm{d}u)\) where \(u = 2x + 1\) | M1 | 3.1a |
| \(= \displaystyle\int \dfrac{1}{4}u^{-2} - \dfrac{1}{4}u^{-3}\,\mathrm{d}u = \ldots\) | dM1 | 1.1b |
| \(= -\dfrac{1}{4}u^{-1} + \dfrac{1}{8}u^{-2}\) | A1 | 1.1b |
| \(\displaystyle\int_0^2 \dfrac{x}{(2x+1)^3}\,\mathrm{d}x = \left[-\dfrac{1}{4}u^{-1} + \dfrac{1}{8}u^{-2}\right]_1^5 = \left(-\dfrac{1}{20} + \dfrac{1}{200}\right) - \left(-\dfrac{1}{4} + \dfrac{1}{8}\right)\) | ddM1 | 1.1b |
| e.g. \(-\dfrac{1}{20} + \dfrac{1}{200} + \dfrac{1}{8} = \dfrac{2}{25}\) * | A1* | 2.1 |
| (5) |
M1: Obtains \(\alpha x(2x+1)^{-2} \pm \beta\displaystyle\int (2x+1)^{-2}\,(\mathrm{d}x)\) o.e. where \(\alpha, \beta \neq 0\) but may be equal to each other (you do not need to be concerned about how they arrive at this)
dM1: Uses a correct method to integrate an expression of the form \(\pm\beta\displaystyle\int (2x+1)^{-2}\,(\mathrm{d}x) \rightarrow \pm\gamma(2x+1)^{-1},\ \ \beta, \gamma \neq 0\)
It is dependent on the previous method mark.
A1: \(-\dfrac{x}{4(2x+1)^2} - \dfrac{1}{8(2x+1)}\) o.e. Allow this to be unsimplified
Watch out for the DI method
| D | I | |
|---|---|---|
| \(+\) | \(x\) | \((2x+1)^{-3}\) |
| \(-\) | \(1\) | \(-\dfrac{(2x+1)^{-2}}{4}\) |
| \(+\) | \(0\) | \(\dfrac{(2x+1)^{-1}}{8}\) |
(arrows in the table go diagonally from \(x\) to \(-\dfrac{(2x+1)^{-2}}{4}\) and from 1 to \(\dfrac{(2x+1)^{-1}}{8}\))
Giving correct integration e.g. \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,\mathrm{d}x = -\dfrac{x(2x+1)^{-2}}{4} - \dfrac{(2x+1)^{-1}}{8}\)
Score M1dM1 for obtaining \(\pm px(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) and A1 for both correct.
ddM1: Substitutes 0 and 2 into an expression of the form \(\pm px(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) or equivalent and subtracts either way round. Evidence of limits used cannot be the given answer. Condone slips but not directly evaluating the expression as 0 when \(x\) is 0. It is dependent on both of the previous method marks.
A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(x\) throughout
Alternative method I – substitution using \(u = 2x + 1\)
M1: Uses a suitable substitution e.g. \(u = 2x + 1\) and proceeds to \(A\displaystyle\int \dfrac{u - 1}{u^3}\,(\mathrm{d}u)\) o.e.
dM1: Splits into separate fractions and attempts to integrate \(A\displaystyle\int u^{-2} - u^{-3}\,(\mathrm{d}u)\) Look for at least one correct index for one of the two terms i.e. \(u^{-2} \rightarrow u^{-1}\) or \(u^{-3} \rightarrow u^{-2}\). It is dependent on the previous method mark.
A1: \(-\dfrac{1}{4}u^{-1} + \dfrac{1}{8}u^{-2}\) o.e.
ddM1: Substitutes correct limits (1 and 5 if in terms of \(u\)) into an expression of the correct form \(\ldots u^{-1} + \ldots u^{-2}\) (or may have substituted back in terms of \(x\) and substitutes in 0 and 2 into an expression of the correct form \(\pm p(2x+1)^{-2} \pm q(2x+1)^{-1},\ \ p, q \neq 0\) o.e and subtracts either way round – see above for ddM1). Evidence of limits used cannot be the given answer. Condone slips. It is dependent on both of the previous method marks.
A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(u\) throughout
Alternative method II – partial fractions
M1: Writes \(\dfrac{x}{(2x+1)^3}\) as \(\dfrac{0.5}{(2x+1)^2} - \dfrac{0.5}{(2x+1)^3}\) o.e. Allow \(\pm\dfrac{M}{(2x+1)^2} \pm \dfrac{N}{(2x+1)^3}\) (where \(M\) and \(N\) are constants)
dM1: \(\displaystyle\int \dfrac{x}{(2x+1)^3}\,(\mathrm{d}x) = \int \dfrac{\text{``}0.5\text{''}}{(2x+1)^2}\,(\mathrm{d}x) - \int \dfrac{\text{``}0.5\text{''}}{(2x+1)^3}\,(\mathrm{d}x) = \pm\ldots(2x+1)^{-1} \pm \ldots(2x+1)^{-2}\)
A1: \(-\dfrac{1}{4(2x+1)^1} + \dfrac{1}{8(2x+1)^2}\) o.e.
ddM1: Substitutes 0 and 2 into an expression of the correct form \(\pm\ldots(2x+1)^{-1} \pm \ldots(2x+1)^{-2}\) and subtracts either way round. Evidence of limits used cannot be the given answer. Condone slips but not directly evaluating the expression as 0 when \(x\) is 0. It is dependent on both of the previous method marks.
A1*: Shows evidence of evaluating after or when the limits are substituted in before proceeding to the given answer with no errors seen including invisible brackets and all previous marks scored. Condone missing/poor notation e.g. missing the d\(x\) throughout.