| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Square root substitution: definite integral |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 techniques: finding intercepts, product rule differentiation, finding turning points, integration by substitution, and product rule with chain rule. Part (ii) requires careful application of the product rule but is routine. Part (iv) involves a standard u-substitution with clear guidance. All parts are textbook-style exercises requiring competent technique but no novel insight, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-1, 0)\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x\sqrt{1 + x} = x(1+x)^{\frac{1}{2}}\) | M1 | |
| \(\Rightarrow \frac{dy}{dx} = 1 \cdot (1+x)^{\frac{1}{2}} + x \cdot \frac{1}{2}(1+x)^{-\frac{1}{2}}\) | A1 | Any correct expression; Combining fractions |
| \(= \frac{2(1+x) + x}{2(1+x)^{\frac{1}{2}}} = \frac{3x + 2}{2\sqrt{1+x}}\) | E1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| At a turning point, gradient is zero. \(x = -\frac{2}{3}\) there. Then \(y = -\frac{2}{3}\sqrt{\frac{1}{3}} = -\frac{2\sqrt{3}}{9}\) | M1, A1 | Accept a simplified form with \(\sqrt{3}\) in the bottom; Accept reference to infinity or to vertical. |
| These are the coordinates of the TP. At P the gradient is undefined. | A1, B1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^0 x\sqrt{1+x}dx = \int_0^1 (u-1)u^{\frac{1}{2}}du\), \(u = 1 + x \Rightarrow \frac{du}{dx} = 1\) | M1 | Integral in \(u\); Change of limits |
| \(= \int_0^1 (u^{\frac{3}{2}} - u^{\frac{1}{2}})du\) | E1 | |
| \(= \left[\frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}}\right]_0^1 = \frac{4}{15}\) | M1, A1 | Integrating. 7 marks |
| The magnitude represents the area bounded by the curve and axis between P and O. It is negative because the curve is below the axis (except at the end points). | B1, B1 | One for geometry and one for sign. |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x\sqrt{1 + x}\sin 2x\) | (given) | |
| \(\Rightarrow \frac{dy}{dx} = \frac{3x+2}{2\sqrt{1+x}}\sin 2x + x\sqrt{1+x} \cdot 2\cos 2x\) | M1, A1 | 2 marks |
### (i)
$P(-1, 0)$ | B1 | 1 mark
### (ii)
$y = x\sqrt{1 + x} = x(1+x)^{\frac{1}{2}}$ | M1
$\Rightarrow \frac{dy}{dx} = 1 \cdot (1+x)^{\frac{1}{2}} + x \cdot \frac{1}{2}(1+x)^{-\frac{1}{2}}$ | A1 | Any correct expression; Combining fractions
$= \frac{2(1+x) + x}{2(1+x)^{\frac{1}{2}}} = \frac{3x + 2}{2\sqrt{1+x}}$ | E1 | 4 marks
### (iii)
At a turning point, gradient is zero. $x = -\frac{2}{3}$ there. Then $y = -\frac{2}{3}\sqrt{\frac{1}{3}} = -\frac{2\sqrt{3}}{9}$ | M1, A1 | Accept a simplified form with $\sqrt{3}$ in the bottom; Accept reference to infinity or to vertical.
These are the coordinates of the TP. At P the gradient is undefined. | A1, B1 | 4 marks
### (iv)
$\int_{-1}^0 x\sqrt{1+x}dx = \int_0^1 (u-1)u^{\frac{1}{2}}du$, $u = 1 + x \Rightarrow \frac{du}{dx} = 1$ | M1 | Integral in $u$; Change of limits
$= \int_0^1 (u^{\frac{3}{2}} - u^{\frac{1}{2}})du$ | E1
$= \left[\frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}}\right]_0^1 = \frac{4}{15}$ | M1, A1 | Integrating. 7 marks
The magnitude represents the area bounded by the curve and axis between P and O. It is negative because the curve is below the axis (except at the end points). | B1, B1 | One for geometry and one for sign.
### (v)
$y = x\sqrt{1 + x}\sin 2x$ | (given)
$\Rightarrow \frac{dy}{dx} = \frac{3x+2}{2\sqrt{1+x}}\sin 2x + x\sqrt{1+x} \cdot 2\cos 2x$ | M1, A1 | 2 marks
---Fig. 8 shows the graph of $y = x\sqrt{1 + x}$. The point P on the curve is on the $x$-axis.
\includegraphics{figure_8}
\begin{enumerate}[label=(\roman*)]
\item Write down the coordinates of P. [1]
\item Show that $\frac{dy}{dx} = \frac{3x + 2}{2\sqrt{1 + x}}$. [4]
\item Hence find the coordinates of the turning point on the curve.
What can you say about the gradient of the curve at P? [4]
\item By using a suitable substitution, show that $\int_0^0 x\sqrt{1 + x} dx = \int_0^1 \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right) du$.
Evaluate this integral, giving your answer in an exact form.
What does this value represent? [7]
\item Use your answer to part (ii) to differentiate $y = x\sqrt{1 + x} \sin 2x$ with respect to $x$.
(You need not simplify your result.) [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q8 [18]}}