| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under polynomial curve |
| Difficulty | Standard +0.3 This is a straightforward calculus problem requiring finding a turning point via differentiation, then integrating a polynomial to find area. All steps are routine A-level techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Overall method of finding the \(x\) coordinate of \(A\) | M1 | Need to see attempt to differentiate with at least two correct terms, and set \(\frac{dy}{dx}=0\) then solve |
| \(y=2x^3-17x^2+40x \Rightarrow \frac{dy}{dx}=6x^2-34x+40\) | B1 | May be unsimplified |
| \(\frac{dy}{dx}=0 \Rightarrow 6x^2-34x+40=0 \Rightarrow 2(3x-5)(x-4)=0 \Rightarrow x=\ldots\) | M1 | Sets \(\frac{dy}{dx}=0\), must be 3TQ in \(x\), attempts to solve via factorisation, formula or calculator |
| Chooses \(x=4\) | A1 | May be awarded from upper limit in their integral. Condone \((x-4)\left(x-\frac{5}{3}\right)\) |
| \(\int 2x^3-17x^2+40x\,dx = \left[\frac{1}{2}x^4 - \frac{17}{3}x^3+20x^2\right]\) | B1 | May be unsimplified |
| Area \(= \frac{1}{2}(4)^4 - \frac{17}{3}(4)^3+20(4)^2\) | M1 | Correct attempt at area with two correct terms. Upper limit must be larger solution of \(\frac{dy}{dx}=0\); lower limit must be 0 |
| \(= \frac{256}{3}\) | A1* | Correct notation and no errors. This is a given answer — must see integration and embedded/calculated values |
# Question 13:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Overall method of finding the $x$ coordinate of $A$ | M1 | Need to see attempt to differentiate with at least **two correct terms**, and set $\frac{dy}{dx}=0$ then solve |
| $y=2x^3-17x^2+40x \Rightarrow \frac{dy}{dx}=6x^2-34x+40$ | B1 | May be unsimplified |
| $\frac{dy}{dx}=0 \Rightarrow 6x^2-34x+40=0 \Rightarrow 2(3x-5)(x-4)=0 \Rightarrow x=\ldots$ | M1 | Sets $\frac{dy}{dx}=0$, must be 3TQ in $x$, attempts to solve via factorisation, formula or calculator |
| Chooses $x=4$ | A1 | May be awarded from upper limit in their integral. Condone $(x-4)\left(x-\frac{5}{3}\right)$ |
| $\int 2x^3-17x^2+40x\,dx = \left[\frac{1}{2}x^4 - \frac{17}{3}x^3+20x^2\right]$ | B1 | May be unsimplified |
| Area $= \frac{1}{2}(4)^4 - \frac{17}{3}(4)^3+20(4)^2$ | M1 | Correct attempt at area with **two correct terms**. Upper limit must be larger solution of $\frac{dy}{dx}=0$; lower limit must be 0 |
| $= \frac{256}{3}$ | A1* | Correct notation and no errors. This is a given answer — must see integration and embedded/calculated values |13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{deba6a2b-1821-4110-bde8-bde18a5f9be9-32_800_787_244_644}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation
$$y = 2 x ^ { 3 } - 17 x ^ { 2 } + 40 x$$
The curve has a minimum turning point at $x = k$.\\
The region $R$, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and the line with equation $x = k$.
Show that the area of $R$ is $\frac { 256 } { 3 }$\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q13 [7]}}