| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find constant from definite integral |
| Difficulty | Moderate -0.8 This is a straightforward AS-level integration question requiring basic power rule integration (including negative powers) and substitution into limits. Part (a) is routine manipulation, and part (b) involves simple algebraic solving for k. No problem-solving insight needed, just direct application of standard techniques. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^n \rightarrow x^{n+1}\) | M1 | For \(x^{-3}\to x^{-2}\) or \(x^1\to x^2\). Implied by sight of \(x^{-2}\) or \(x^2\) |
| \(\int\left(\frac{4}{x^3}+kx\right)dx = -\frac{2}{x^2}+\frac{1}{2}kx^2\) | A1 A1 | First A1: either term correct (unsimplified). Second A1: correct and simplified with \(+c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left[-\frac{2}{x^2}+\frac{1}{2}kx^2\right]^2_{0.5} = \left(-\frac{2}{2^2}+\frac{1}{2}k\times4\right)-\left(-\frac{2}{(0.5)^2}+\frac{1}{2}k\times(0.5)^2\right)=8\) | M1 | Substitutes both limits into \(-\frac{2}{x^2}+\frac{1}{2}kx^2\), subtracts either way and sets equal to 8 |
| \(7.5+\frac{15}{8}k=8\) | dM1 | Solving a linear equation in \(k\); dependent on previous M |
| \(k=\frac{4}{15}\) oe | A1 | Accept \(\frac{m}{n}\) where \(\frac{m}{n}=\frac{4}{15}\). Condone recurring decimal \(0.\dot{6}\) but not 0.266 or 0.267 |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^n \rightarrow x^{n+1}$ | M1 | For $x^{-3}\to x^{-2}$ or $x^1\to x^2$. Implied by sight of $x^{-2}$ or $x^2$ |
| $\int\left(\frac{4}{x^3}+kx\right)dx = -\frac{2}{x^2}+\frac{1}{2}kx^2$ | A1 A1 | First A1: either term correct (unsimplified). Second A1: correct and simplified with $+c$ |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left[-\frac{2}{x^2}+\frac{1}{2}kx^2\right]^2_{0.5} = \left(-\frac{2}{2^2}+\frac{1}{2}k\times4\right)-\left(-\frac{2}{(0.5)^2}+\frac{1}{2}k\times(0.5)^2\right)=8$ | M1 | Substitutes both limits into $-\frac{2}{x^2}+\frac{1}{2}kx^2$, subtracts either way and sets equal to 8 |
| $7.5+\frac{15}{8}k=8$ | dM1 | Solving a **linear** equation in $k$; dependent on previous M |
| $k=\frac{4}{15}$ oe | A1 | Accept $\frac{m}{n}$ where $\frac{m}{n}=\frac{4}{15}$. Condone recurring decimal $0.\dot{6}$ but not 0.266 or 0.267 |
---\begin{enumerate}
\item (a) Given that $k$ is a constant, find
\end{enumerate}
$$\int \left( \frac { 4 } { x ^ { 3 } } + k x \right) \mathrm { d } x$$
simplifying your answer.\\
(b) Hence find the value of $k$ such that
$$\int _ { 0.5 } ^ { 2 } \left( \frac { 4 } { x ^ { 3 } } + k x \right) \mathrm { d } x = 8$$
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q3 [6]}}