| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.8 This is a challenging C4 integration by parts question requiring algebraic division, recognition of the integration by parts structure with logarithm, and careful algebraic manipulation to reach the specified form. The multi-step nature, need to connect part (i) to part (ii), and the non-routine algebraic simplification at the end elevate this above a standard integration by parts exercise, though it remains within the scope of a well-prepared C4 student. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t^2\) in quotient and \(t^3 + 2t^2\) seen | B1 | or \(\dfrac{t(t^2-4)+4t}{(t+2)}\) |
| \(-2t\) in quotient and \(-2t^2 - (-2t^2 - 4t) = 4t\) seen | B1 | \(\dfrac{t(t+2)(t-2)}{(t+2)} + \dfrac{4t}{t+2}\) |
| completion to obtain correct quotient and remainder identified | B1 | \(t(t-2) + \dfrac{4(t+2)-8}{t+2}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{t^3}{t+2} \equiv At^2 + Bt + C + \dfrac{D}{(t+2)}\) | B1 | or \(t^3 \equiv (At^2 + Bt + C)(t+2) + D\) |
| equate coefficients to obtain correctly \(A=1\), \(0=2A+B\) and \(B=-2\) | B1 | |
| \(0 = 2B + C\) and \(0 = 2C + D\) obtained and solved correctly | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| integration by parts with \(u = \ln(t+2)\) and \(dv = 6t^2\) to obtain \(f(t) \pm \int g(t)\,dt\) | M1* | \(f(t)\) must include \(t^3\) and \(g(t)\) must not include a logarithm |
| \(2t^3\ln(t+2) - \int\dfrac{2t^3}{t+2}\,dt\) cao | A1 | |
| result from part (i) seen in integrand; must follow award of at least first M1 | M1* | no integration required for this mark |
| \(F[t] = 2t^3\ln(t+2) \pm \dfrac{2t^3}{3} \pm 2t^2 \pm 8t \pm 16\ln(t+2)\) | A1 | \(2t^3\ln(t+2) - \dfrac{2t^3}{3} + 2t^2 - 8t + 16\ln(t+2)\) |
| their \(F[2] - F[1]\) | M1dep* | at least one of their terms correctly integrated |
| \(-6\frac{2}{3} - 18\ln 3 + 32\ln 4\) oe cao | A1 | |
| [6] |
# Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t^2$ in quotient and $t^3 + 2t^2$ seen | B1 | or $\dfrac{t(t^2-4)+4t}{(t+2)}$ | or $\dfrac{(t+2)^3 - 6t^2 - 12t - 8}{(t+2)}$ |
| $-2t$ in quotient and $-2t^2 - (-2t^2 - 4t) = 4t$ seen | B1 | $\dfrac{t(t+2)(t-2)}{(t+2)} + \dfrac{4t}{t+2}$ | $\dfrac{(t+2)^3}{(t+2)} - \dfrac{6((t+2)^2 - 4t - 4) + 12t + 8}{(t+2)}$ oe |
| completion to obtain correct quotient and remainder identified | B1 | $t(t-2) + \dfrac{4(t+2)-8}{t+2}$ | $(t+2)^2 - 6(t+2) + \dfrac{12t+16}{t+2}$ oe; both steps needed for final B1 |
| **[3]** | | |
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# Question 8(i) Alternative:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{t^3}{t+2} \equiv At^2 + Bt + C + \dfrac{D}{(t+2)}$ | B1 | or $t^3 \equiv (At^2 + Bt + C)(t+2) + D$ | or B1 for $\dfrac{t^2(t+2) - 2t^2}{(t+2)}$ |
| equate coefficients to obtain correctly $A=1$, $0=2A+B$ and $B=-2$ | B1 | | B1 for $t^2 + \dfrac{-2t(t+2)+4t}{(t+2)}$ |
| $0 = 2B + C$ and $0 = 2C + D$ obtained and solved correctly | B1 | | B1 for $t^2 - 2t + \dfrac{4(t+2)-8}{(t+2)}$ |
| **[3]** | | |
---
# Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| integration by parts with $u = \ln(t+2)$ and $dv = 6t^2$ to obtain $f(t) \pm \int g(t)\,dt$ | M1* | $f(t)$ must include $t^3$ and $g(t)$ must **not** include a logarithm | ignore spurious $dx$ etc |
| $2t^3\ln(t+2) - \int\dfrac{2t^3}{t+2}\,dt$ cao | A1 | | alternatively, following $u = t+2$: $\int 2(u^2 - 6u + 12 - \dfrac{8}{u})\,du$ oe |
| result from part (i) seen in integrand; must follow award of at least first M1 | M1* | no integration required for this mark | |
| $F[t] = 2t^3\ln(t+2) \pm \dfrac{2t^3}{3} \pm 2t^2 \pm 8t \pm 16\ln(t+2)$ | A1 | $2t^3\ln(t+2) - \dfrac{2t^3}{3} + 2t^2 - 8t + 16\ln(t+2)$ | $\dfrac{2u^3}{3} - 6u^2 + 24u - 16\ln u$ and $2t^3\ln(t+2)$ |
| their $F[2] - F[1]$ | M1dep* | at least one of their terms correctly integrated | NB limits following substitution are $u=4$ and $u=3$ |
| $-6\frac{2}{3} - 18\ln 3 + 32\ln 4$ oe cao | A1 | |
| **[6]** | | |
---8 (i) Use division to show that $\frac { t ^ { 3 } } { t + 2 } \equiv t ^ { 2 } - 2 t + 4 - \frac { 8 } { t + 2 }$.\\
(ii) Find $\int _ { 1 } ^ { 2 } 6 t ^ { 2 } \ln ( t + 2 ) \mathrm { d } t$. Give your answer in the form $A + B \ln 3 + C \ln 4$.
\hfill \mbox{\textit{OCR C4 2014 Q8 [9]}}