| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Moderate -0.8 This is a straightforward C1 integration question requiring standard techniques: direct integration of a polynomial, finding area between curve and horizontal line using subtraction, and finding tangent equation via differentiation. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^2 (x^4 - 8x + 9)\,dx = \left[\frac{x^5}{5} - 4x^2 + 9x\right]_0^2\); \(= \frac{32}{5} - 16 + 18 = \frac{32}{5} + 2 = \frac{42}{5}\) | M1 A1 A1 M1 A1 | M1 for integration attempt; A1A1 for correct terms; M1 for substituting limits; A1 for \(\frac{42}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Area of rectangle \(= 2 \times 9 = 18\); shaded area \(= 18 - \frac{42}{5} = \frac{48}{5}\) | M1 A1 | M1 for \(18 - \frac{42}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4x^3 - 8\); at \(x=1\): \(\frac{dy}{dx} = 4-8 = -4\) | M1 A1 M1 A1 | M1 for differentiation; M1 for substituting \(x=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 2 = -4(x-1)\), i.e. \(y = -4x + 6\) | B1 | Follow through on gradient |
# Question 4:
## Part (a)(i)
| $\int_0^2 (x^4 - 8x + 9)\,dx = \left[\frac{x^5}{5} - 4x^2 + 9x\right]_0^2$; $= \frac{32}{5} - 16 + 18 = \frac{32}{5} + 2 = \frac{42}{5}$ | M1 A1 A1 M1 A1 | M1 for integration attempt; A1A1 for correct terms; M1 for substituting limits; A1 for $\frac{42}{5}$ |
## Part (a)(ii)
| Area of rectangle $= 2 \times 9 = 18$; shaded area $= 18 - \frac{42}{5} = \frac{48}{5}$ | M1 A1 | M1 for $18 - \frac{42}{5}$ |
## Part (b)(i)
| $\frac{dy}{dx} = 4x^3 - 8$; at $x=1$: $\frac{dy}{dx} = 4-8 = -4$ | M1 A1 M1 A1 | M1 for differentiation; M1 for substituting $x=1$ |
## Part (b)(ii)
| $y - 2 = -4(x-1)$, i.e. $y = -4x + 6$ | B1 | Follow through on gradient |
---4 The curve with equation $y = x ^ { 4 } - 8 x + 9$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-5_410_609_383_721}
The point $( 2,9 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int _ { 0 } ^ { 2 } \left( x ^ { 4 } - 8 x + 9 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve and the line $y = 9$.
\end{enumerate}\item The point $A ( 1,2 )$ lies on the curve with equation $y = x ^ { 4 } - 8 x + 9$.
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of the curve at the point $A$.
\item Hence find an equation of the tangent to the curve at the point $A$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q4 [12]}}