| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring basic polynomial integration and application of boundary conditions. Part (a) involves integrating x² - 3x and finding the constant using the given point - a routine C1/C2 exercise. Part (b) requires substituting the found equation and integrating again, which is mechanical calculation with no conceptual challenge. Both parts are standard textbook exercises with no problem-solving insight required. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int (x^2 - 3x)\,dx\) | M1 | \(\geq\) one term or both powers correct. May be implied by result |
| \(= \frac{x^3}{3} - \frac{3x^2}{2} + c\) | A1 | Allow without "\(+c\)" |
| \(20 = \frac{6^3}{3} - \frac{3\times6^2}{2} + c \quad (\Rightarrow c = 2)\) | M1 | Substitute \(x=6\) into their integral, dep M1, \(\&=20\) |
| \(y = \frac{x^3}{3} - \frac{3x^2}{2} + 2\) | A1 | Correct answer, including "\(y=\)". Allow \(f(x) = \ldots\) NB: if no working seen for finding \(c\), but fully correct answer given: SC3 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_1^p \!\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2\right)dx\) | M1 | ft their equation, dep cubic. \(\geq\) two terms or all three powers correct. May be implied by result |
| \(= \left[\frac{x^4}{12} - \frac{x^3}{2} + 2x\right]_1^p\) | A1ft | Correct integral of their curve, dep quartic |
| Substitute limits 1 and \(p\) | M1 | dep integration attempted |
| \(= \frac{p^4}{12} - \frac{p^3}{2} + 2p - \frac{19}{12}\) oe | A1ft | ft their integral, dep their integral is a quartic |
| [4] |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int (x^2 - 3x)\,dx$ | M1 | $\geq$ one term or both powers correct. May be implied by result |
| $= \frac{x^3}{3} - \frac{3x^2}{2} + c$ | A1 | Allow without "$+c$" |
| $20 = \frac{6^3}{3} - \frac{3\times6^2}{2} + c \quad (\Rightarrow c = 2)$ | M1 | Substitute $x=6$ into their integral, dep M1, $\&=20$ |
| $y = \frac{x^3}{3} - \frac{3x^2}{2} + 2$ | A1 | Correct answer, including "$y=$". Allow $f(x) = \ldots$ NB: if no working seen for finding $c$, but fully correct answer given: SC3 |
| **[4]** | | |
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## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^p \!\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2\right)dx$ | M1 | ft their equation, dep cubic. $\geq$ two terms or all three powers correct. May be implied by result |
| $= \left[\frac{x^4}{12} - \frac{x^3}{2} + 2x\right]_1^p$ | A1ft | Correct integral of their curve, dep quartic |
| Substitute limits 1 and $p$ | M1 | dep integration attempted |
| $= \frac{p^4}{12} - \frac{p^3}{2} + 2p - \frac{19}{12}$ oe | A1ft | ft their integral, dep their integral is a quartic |
| **[4]** | | |
---5 The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } - 3 x$. The curve passes through the point (6, 20).
\begin{enumerate}[label=(\alph*)]
\item Determine the equation of the curve.
\item Hence determine $\int _ { 1 } ^ { p } y \mathrm {~d} x$ in terms of the constant $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [8]}}