| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Find curve from gradient |
| Difficulty | Moderate -0.8 This is a straightforward C2 integration question testing basic techniques: expanding and integrating a polynomial, integrating a power of x (x^{-1/2}), and finding a constant of integration using a boundary condition. All parts are routine textbook exercises requiring only direct application of standard rules with no problem-solving or insight needed. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(x^3+2x)\,dx = \frac{1}{4}x^4 + x^2 + c\) | M1 | For expanding and integration attempt |
| A1 | For \(\frac{1}{4}x^4 + x^2\) correct | |
| B1 | 3 For addition of an arbitrary constant (this mark can be given in (b)(i) if not earned here), and no \(dx\) in either |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x^{-\frac{1}{2}}\,dx = 2x^{\frac{1}{2}} + c\) | B1 | For use of \(\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\) |
| M1 | For integral of the form \(kx^{\frac{1}{2}}\) | |
| A1 | 3 For correct term \(2x^{\frac{1}{2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = 2\sqrt{4} + c \Rightarrow c = -4\) | M1 | For use of \(x=4\), \(y=0\) to evaluate \(c\) |
| A1ft | For correct \(c\) from their answer in (b)(i) | |
| Hence curve is \(y = 2x^{\frac{1}{2}} - 4\) | A1ft | 3 For equation of the curve correctly stated |
# Question 6:
## Part (a):
| $\int(x^3+2x)\,dx = \frac{1}{4}x^4 + x^2 + c$ | M1 | For expanding and integration attempt |
|---|---|---|
| | A1 | For $\frac{1}{4}x^4 + x^2$ correct |
| | B1 | **3** For addition of an arbitrary constant (this mark can be given in **(b)(i)** if not earned here), and no $dx$ in either |
## Part (b)(i):
| $\int x^{-\frac{1}{2}}\,dx = 2x^{\frac{1}{2}} + c$ | B1 | For use of $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$ |
|---|---|---|
| | M1 | For integral of the form $kx^{\frac{1}{2}}$ |
| | A1 | **3** For correct term $2x^{\frac{1}{2}}$ |
## Part (b)(ii):
| $0 = 2\sqrt{4} + c \Rightarrow c = -4$ | M1 | For use of $x=4$, $y=0$ to evaluate $c$ |
|---|---|---|
| | A1ft | For correct $c$ from their answer in **(b)(i)** |
| Hence curve is $y = 2x^{\frac{1}{2}} - 4$ | A1ft | **3** For equation of the curve correctly stated |
---6
\begin{enumerate}[label=(\alph*)]
\item Find $\int x \left( x ^ { 2 } + 2 \right) \mathrm { d } x$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \frac { 1 } { \sqrt { x } } \mathrm {~d} x$.
\item The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { } x }$. Find the equation of the curve, given that it passes through the point $( 4,0 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR C2 2005 Q6 [9]}}