| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a standard C2 calculus question requiring differentiation of a power function (rewriting √x as x^(1/2)), finding a tangent equation using point-slope form, then calculating an area using integration. All techniques are routine for C2 level with clear structure and no novel problem-solving required. The 14 marks reflect length rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = 4\) \(\therefore y = 12 - 8 + 2 = 6\) | B1 | |
| \(\frac{dy}{dx} = 3 - 2x^{-1}\) | M1 A1 | |
| \(\text{grad} = 3 - 1 = 2\) | M1 | |
| \(\therefore y - 6 = 2(x - 4)\) | M1 | |
| \(y = 2x - 2\) | A1 | |
| (b) area under curve \(= \int_0^4 (3x - 4\sqrt{x} + 2) \, dx\) | ||
| \(= [\frac{3}{2}x^2 - \frac{8}{3}x^{\frac{3}{2}} + 2x]_0^4\) | M1 A2 | |
| \(= (24 - \frac{64}{3} + 8) - (0) = 10\frac{2}{3}\) | M1 | |
| tangent meets x-axis when \(y = 0 \Rightarrow x = 1\) | M1 | |
| area of triangle \(= \frac{1}{2} \times 3 \times 6 = 9\) | A1 | |
| shaded area \(= 10\frac{2}{3} - 9 = \frac{5}{3}\) | M1 A1 | (14) |
**(a)** $x = 4$ $\therefore y = 12 - 8 + 2 = 6$ | B1 |
$\frac{dy}{dx} = 3 - 2x^{-1}$ | M1 A1 |
$\text{grad} = 3 - 1 = 2$ | M1 |
$\therefore y - 6 = 2(x - 4)$ | M1 |
$y = 2x - 2$ | A1 |
**(b)** area under curve $= \int_0^4 (3x - 4\sqrt{x} + 2) \, dx$ | |
$= [\frac{3}{2}x^2 - \frac{8}{3}x^{\frac{3}{2}} + 2x]_0^4$ | M1 A2 |
$= (24 - \frac{64}{3} + 8) - (0) = 10\frac{2}{3}$ | M1 |
tangent meets x-axis when $y = 0 \Rightarrow x = 1$ | M1 |
area of triangle $= \frac{1}{2} \times 3 \times 6 = 9$ | A1 |
shaded area $= 10\frac{2}{3} - 9 = \frac{5}{3}$ | M1 A1 | (14)\includegraphics{figure_2}
Figure 2 shows the curve $C$ with equation $y = 3x - 4\sqrt{x} + 2$ and the tangent to $C$ at the point $A$.
Given that $A$ has $x$-coordinate 4,
\begin{enumerate}[label=(\alph*)]
\item show that the tangent to $C$ at $A$ has the equation $y = 2x - 2$. [6]
\end{enumerate}
The shaded region is bounded by $C$, the tangent to $C$ at $A$ and the positive coordinate axes.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the area of the shaded region. [8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [14]}}