| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Moderate -0.3 This is a standard A-level integration question requiring routine application of definite integrals to find areas. Part (a) involves straightforward integration of a cubic between roots, part (b) requires setting up an equation and algebraic verification (not solving), and part (c) tests interpretation of roots. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x(x+2)(x-4) = x^3 - 2x^2 - 8x\) | B1 | Expands correctly; may be in different order |
| \(\int x^3 - 2x^2 - 8x \, dx \rightarrow \frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2\) | M1 | Correct attempt at integration; look for \(x^n \rightarrow x^{n+1}\) seen at least twice |
| Attempts area using correct strategy \(\int_{-2}^{0} y \, dx\) | dM1 | Dependent on previous M; requires substitution of \(-2\) into \(\pm\) their integrated function |
| \(\left[\frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2\right]_{-2}^{0} = (0) - \left(4 - \frac{-16}{3} - 16\right) = \frac{20}{3}\) * | A1* | Limits must be correct way around; embedded or calculated values must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For setting \(\frac{1}{4}b^4 - \frac{2}{3}b^3 - 4b^2 = \pm\frac{20}{3}\) | M1 | Must have followed \(\frac{1}{4}b^4 - \frac{2}{3}b^3 - 4b^2 = -\frac{20}{3}\) |
| Correctly deduces \(3b^4 - 8b^3 - 48b^2 + 80 = 0\) | A1 | Terms may be in different order; must have integer coefficients |
| Attempts to factorise: \(3b^4 - 8b^3 - 48b^2 \pm 80 = (b+2)(b+2)(3b^2\ldots b\ldots 20)\) | M1 | Via repeated division or inspection |
| Achieves \((b+2)^2(3b^2 - 20b + 20) = 0\) with no errors | A1* | Must equal 0; alternatively attempts to expand achieving quartic expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States that between \(x = -2\) and \(x = 5.442\) the area above the \(x\)-axis = area below the \(x\)-axis | B1 | Sketches curve with vertical line to right of 4; \(x = 5.442\) may not be labelled |
| Explanation that area above \(x\)-axis = area below \(x\)-axis with appropriate areas shaded/labelled | B1 | Alternatively states area between 1.225 and 4 equals area between 4 and 5.442; or net area between 0 and 5.442 is \(-\frac{20}{3}\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x(x+2)(x-4) = x^3 - 2x^2 - 8x$ | B1 | Expands correctly; may be in different order |
| $\int x^3 - 2x^2 - 8x \, dx \rightarrow \frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2$ | M1 | Correct attempt at integration; look for $x^n \rightarrow x^{n+1}$ seen at least twice |
| Attempts area using correct strategy $\int_{-2}^{0} y \, dx$ | dM1 | Dependent on previous M; requires substitution of $-2$ into $\pm$ their integrated function |
| $\left[\frac{1}{4}x^4 - \frac{2}{3}x^3 - 4x^2\right]_{-2}^{0} = (0) - \left(4 - \frac{-16}{3} - 16\right) = \frac{20}{3}$ * | A1* | Limits must be correct way around; embedded or calculated values must be seen |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For setting $\frac{1}{4}b^4 - \frac{2}{3}b^3 - 4b^2 = \pm\frac{20}{3}$ | M1 | Must have followed $\frac{1}{4}b^4 - \frac{2}{3}b^3 - 4b^2 = -\frac{20}{3}$ |
| Correctly deduces $3b^4 - 8b^3 - 48b^2 + 80 = 0$ | A1 | Terms may be in different order; must have integer coefficients |
| Attempts to factorise: $3b^4 - 8b^3 - 48b^2 \pm 80 = (b+2)(b+2)(3b^2\ldots b\ldots 20)$ | M1 | Via repeated division or inspection |
| Achieves $(b+2)^2(3b^2 - 20b + 20) = 0$ with no errors | A1* | Must equal 0; alternatively attempts to expand achieving quartic expression |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States that between $x = -2$ and $x = 5.442$ the area above the $x$-axis = area below the $x$-axis | B1 | Sketches curve with vertical line to right of 4; $x = 5.442$ may not be labelled |
| Explanation that area above $x$-axis = area below $x$-axis with appropriate areas shaded/labelled | B1 | Alternatively states area between 1.225 and 4 equals area between 4 and 5.442; or net area between 0 and 5.442 is $-\frac{20}{3}$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-22_812_958_244_555}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = x ( x + 2 ) ( x - 4 )$.\\
The region $R _ { 1 }$ shown shaded in Figure 2 is bounded by the curve and the negative $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the exact area of $R _ { 1 }$ is $\frac { 20 } { 3 }$
The region $R _ { 2 }$ also shown shaded in Figure 2 is bounded by the curve, the positive $x$-axis and the line with equation $x = b$, where $b$ is a positive constant and $0 < b < 4$
Given that the area of $R _ { 1 }$ is equal to the area of $R _ { 2 }$
\item verify that $b$ satisfies the equation
$$( b + 2 ) ^ { 2 } \left( 3 b ^ { 2 } - 20 b + 20 \right) = 0$$
The roots of the equation $3 b ^ { 2 } - 20 b + 20 = 0$ are 1.225 and 5.442 to 3 decimal places. The value of $b$ is therefore 1.225 to 3 decimal places.
\item Explain, with the aid of a diagram, the significance of the root 5.442
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q8 [10]}}