| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Independent multi-part (different techniques) |
| Difficulty | Standard +0.3 Part (i) is a standard two-application integration by parts with e^x (textbook exercise). Part (ii) is a routine substitution leading to polynomial integration, though finding the specific form requires careful algebra. Both are well-practiced techniques at this level with no novel insight required, making this slightly easier than average. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int x^2e^x dx = x^2e^x - \int 2xe^x dx\ (+c)\) | M1A1 | M1 for applying parts to obtain \(x^2e^x - k\int xe^x dx\ (+c)\) where \(k>0\) |
| \(\int xe^x dx = xe^x - \int e^x dx\ (+c)\) | M1 | Applies parts again; obtains \(xe^x - \alpha\int e^x dx\ (+c)\) where \(\alpha>0\) |
| \(\int x^2e^x dx = x^2e^x - 2xe^x + 2e^x\ (+c)\) | A1 | Correct expression; no requirement for \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u=(1-3x)^{\frac{1}{2}} \Rightarrow u^2=1-3x \Rightarrow 2u\frac{du}{dx}=-3\) or \(\frac{du}{dx}=-\frac{3}{2}(1-3x)^{-\frac{1}{2}}\) | B1 | Any correct expression involving \(\frac{du}{dx}\) |
| \(\int\frac{27x}{\sqrt{1-3x}}dx = 27\int\frac{\frac{1-u^2}{3}}{u}\left(-\frac{2u}{3}\right)du\) | M1 | Substitutes \(x=f(u^2)\) to fully change integrand; \(u\)'s may cancel; condone \(du\) not present but cannot appear as \(dx\) |
| \(6\int(u^2-1)du\) or \(-6\int(1-u^2)du\) | A1 | Correct simplified integral; look for \(u\)'s cancelled and simplification of constants |
| \(6\left(\frac{u^3}{3}-u\right)(+k)\) | A1ft | Correct follow through integration |
| \(6\left(\frac{(1-3x)^{\frac{3}{2}}}{3}-(1-3x)^{\frac{1}{2}}\right)(+k) = 2(1-3x)^{\frac{1}{2}}(1-3x-3)(+k)\) | M1 | Achieves \(Pu^3+Qu\) and correctly takes out common factor of \(u\) or \((1-3x)^{\frac{1}{2}}\), substitutes back |
| \(=-2(1-3x)^{\frac{1}{2}}(2+3x)(+k)\) or \(=-2(1-3x)^{\frac{1}{2}}(3x+2)(+k)\) | A1 | CAO; no requirement for \(+k\) |
# Question 5:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^2e^x dx = x^2e^x - \int 2xe^x dx\ (+c)$ | M1A1 | M1 for applying parts to obtain $x^2e^x - k\int xe^x dx\ (+c)$ where $k>0$ |
| $\int xe^x dx = xe^x - \int e^x dx\ (+c)$ | M1 | Applies parts again; obtains $xe^x - \alpha\int e^x dx\ (+c)$ where $\alpha>0$ |
| $\int x^2e^x dx = x^2e^x - 2xe^x + 2e^x\ (+c)$ | A1 | Correct expression; no requirement for $+c$ |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=(1-3x)^{\frac{1}{2}} \Rightarrow u^2=1-3x \Rightarrow 2u\frac{du}{dx}=-3$ or $\frac{du}{dx}=-\frac{3}{2}(1-3x)^{-\frac{1}{2}}$ | B1 | Any correct expression involving $\frac{du}{dx}$ |
| $\int\frac{27x}{\sqrt{1-3x}}dx = 27\int\frac{\frac{1-u^2}{3}}{u}\left(-\frac{2u}{3}\right)du$ | M1 | Substitutes $x=f(u^2)$ to fully change integrand; $u$'s may cancel; condone $du$ not present but cannot appear as $dx$ |
| $6\int(u^2-1)du$ or $-6\int(1-u^2)du$ | A1 | Correct simplified integral; look for $u$'s cancelled and simplification of constants |
| $6\left(\frac{u^3}{3}-u\right)(+k)$ | A1ft | Correct follow through integration |
| $6\left(\frac{(1-3x)^{\frac{3}{2}}}{3}-(1-3x)^{\frac{1}{2}}\right)(+k) = 2(1-3x)^{\frac{1}{2}}(1-3x-3)(+k)$ | M1 | Achieves $Pu^3+Qu$ and correctly takes out common factor of $u$ or $(1-3x)^{\frac{1}{2}}$, substitutes back |
| $=-2(1-3x)^{\frac{1}{2}}(2+3x)(+k)$ or $=-2(1-3x)^{\frac{1}{2}}(3x+2)(+k)$ | A1 | CAO; no requirement for $+k$ |\begin{enumerate}
\item (i) Find
\end{enumerate}
$$\int x ^ { 2 } \mathrm { e } ^ { x } \mathrm {~d} x$$
(4)\\
(ii) Use the substitution $u = \sqrt { 1 - 3 x }$ to show that
$$\int \frac { 27 x } { \sqrt { 1 - 3 x } } \mathrm {~d} x = - 2 ( 1 - 3 x ) ^ { \frac { 1 } { 2 } } ( A x + B ) + k$$
where $A$ and $B$ are integers to be found and $k$ is an arbitrary constant.
\hfill \mbox{\textit{Edexcel P4 2023 Q5 [10]}}