| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Area under curve requiring parts |
| Difficulty | Standard +0.8 This question requires integration by parts applied twice to evaluate ∫₀¹ 8x²e^(-3x)dx, which is a standard but non-trivial technique. Students must correctly apply the method systematically, handle the exponential and polynomial terms carefully, and simplify to the required exact form. While the technique is well-practiced at A-level Further Maths, the two applications and algebraic manipulation make it moderately challenging, placing it above average difficulty. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 8x^2 e^{-3x}\,dx = -\frac{8x^2}{3}e^{-3x} + \int \frac{16x}{3}e^{-3x}\,dx\) | M1, A1 | M1: Obtains \(\pm\alpha x^2 e^{-3x} \pm \beta \int x\,e^{-3x}\,dx\). A1: Correct expression simplified or unsimplified; condone "8" missing |
| \(= -\frac{8x^2}{3}e^{-3x} - \frac{16x}{9}e^{-3x} + \int \frac{16}{9}e^{-3x}\,dx\) | dM1 | Attempts parts again on \(\pm\beta \int x\,e^{-3x}\,dx\) to obtain \(\pm Axe^{-3x} \pm B\int e^{-3x}\,dx\). Depends on first M mark |
| \(\left[-\frac{8x^2}{3}e^{-3x} - \frac{16x}{9}e^{-3x} - \frac{16}{27}e^{-3x}\right]_0^1\) | M1 | Substitutes limits 1 and 0 into expression of form \(\pm\alpha x^2 e^{-3x} \pm \beta xe^{-3x} \pm \gamma e^{-3x}\); must show use of both limits and use of \(e^0=1\) |
| \(= -\frac{8}{3}e^{-3} - \frac{16}{9}e^{-3} - \frac{16}{27}e^{-3} - \left(-0 - 0 - \frac{16}{27}\right)\) | ||
| \(= \frac{16}{27} - \frac{136}{27}e^{-3}\) | A1 | Correct answer; allow equivalent exact fractions; condone \(\frac{16}{27} - \frac{136}{27e^3}\) |
| (5 marks) |
# Question 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 8x^2 e^{-3x}\,dx = -\frac{8x^2}{3}e^{-3x} + \int \frac{16x}{3}e^{-3x}\,dx$ | M1, A1 | M1: Obtains $\pm\alpha x^2 e^{-3x} \pm \beta \int x\,e^{-3x}\,dx$. A1: Correct expression simplified or unsimplified; condone "8" missing |
| $= -\frac{8x^2}{3}e^{-3x} - \frac{16x}{9}e^{-3x} + \int \frac{16}{9}e^{-3x}\,dx$ | dM1 | Attempts parts again on $\pm\beta \int x\,e^{-3x}\,dx$ to obtain $\pm Axe^{-3x} \pm B\int e^{-3x}\,dx$. Depends on first M mark |
| $\left[-\frac{8x^2}{3}e^{-3x} - \frac{16x}{9}e^{-3x} - \frac{16}{27}e^{-3x}\right]_0^1$ | M1 | Substitutes limits 1 and 0 into expression of form $\pm\alpha x^2 e^{-3x} \pm \beta xe^{-3x} \pm \gamma e^{-3x}$; must show use of **both** limits and use of $e^0=1$ |
| $= -\frac{8}{3}e^{-3} - \frac{16}{9}e^{-3} - \frac{16}{27}e^{-3} - \left(-0 - 0 - \frac{16}{27}\right)$ | | |
| $= \frac{16}{27} - \frac{136}{27}e^{-3}$ | A1 | Correct answer; allow equivalent exact fractions; condone $\frac{16}{27} - \frac{136}{27e^3}$ |
| **(5 marks)** | | |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-28_668_743_251_662}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 5 shows a sketch of part of the curve $C$ with equation
$$y = 8 x ^ { 2 } \mathrm { e } ^ { - 3 x } \quad x \geqslant 0$$
The finite region $R$, shown shaded in Figure 5, is bounded by
\begin{itemize}
\item the curve $C$
\item the line with equation $x = 1$
\item the $x$-axis
\end{itemize}
Find the exact area of $R$, giving your answer in the form
$$A + B \mathrm { e } ^ { - 3 }$$
where $A$ and $B$ are rational numbers to be found.
\hfill \mbox{\textit{Edexcel Paper 2 2024 Q11 [5]}}