| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: verifying coordinates, quotient rule differentiation, finding turning points, and integration by substitution. While part (iv) requires careful handling of the substitution and limits, each step follows routine procedures with clear guidance. The substitution is explicitly given, and the question scaffolds the solution methodically. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = -1\), \(y = -1\sqrt{0} = 0\) | E1 | |
| Domain \(x \geq -1\) | B1 [2] | Not \(y \geq -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = x \cdot \frac{1}{2}(1+x)^{-1/2} + (1+x)^{1/2}\) | B1 | \(x \cdot \frac{1}{2}(1+x)^{-1/2}\) |
| \(= \frac{1}{2}(1+x)^{-1/2}[x + 2(1+x)]\) | B1, M1 | \(\ldots + (1+x)^{1/2}\); taking out common factor or common denominator |
| \(= \dfrac{2+3x}{2\sqrt{1+x}}\) * | E1 [4] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = x+1 \Rightarrow du/dx = 1\), \(y = (u-1)u^{1/2} = u^{3/2} - u^{1/2}\) | M1 | |
| \(\frac{dy}{du} = \frac{3}{2}u^{\frac{1}{2}} - \frac{1}{2}u^{-\frac{1}{2}}\) | A1 | |
| \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{3}{2}(x+1)^{\frac{1}{2}} - \frac{1}{2}(x+1)^{-\frac{1}{2}}\) | M1 | |
| \(= \frac{1}{2}(x+1)^{-\frac{1}{2}}(3x+3-1)\) | M1 | taking out common factor or common denominator |
| \(= \dfrac{2+3x}{2\sqrt{1+x}}\) * | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(dy/dx = 0\) when \(3x + 2 = 0\) | M1, A1cao | |
| \(x = -\frac{2}{3}\) | o. | |
| \(y = -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}\) | A1 | not \(x \geq -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}\) (ft their \(y\) value, even if approximate) |
| Range is \(y \geq -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}\) | B1 ft [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_{-1}^{0} x\sqrt{1+x}\, dx\) | ||
| Let \(u = 1+x\), \(du/dx = 1 \Rightarrow du = dx\) | M1 | \(du = dx\) or \(du/dx = 1\) or \(dx/du = 1\) |
| When \(x=-1\), \(u=0\); when \(x=0\), \(u=1\) | B1 | changing limits – allow with no working shown provided limits are present and consistent with \(dx\) and \(du\) |
| \(= \int_{0}^{1}(u-1)\sqrt{u}\, du\) | M1 | \((u-1)\sqrt{u}\) |
| \(= \int_{0}^{1}(u^{3/2} - u^{1/2})\, du\) * | E1 | www – condone only final brackets missing, otherwise notation must be correct |
| \(= \left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_{0}^{1}\) | B1 B1 | \(\frac{2}{5}u^{5/2}\), \(-\frac{2}{3}u^{3/2}\) (oe) |
| substituting correct limits | M1 | (can imply the zero limit) |
| \(= \pm\dfrac{4}{15}\) | A1cao [8] | \(\pm\frac{4}{15}\) or \(\pm 0.27\) or better, not \(0.26\) |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = -1$, $y = -1\sqrt{0} = 0$ | E1 | |
| Domain $x \geq -1$ | B1 [2] | Not $y \geq -1$ |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x \cdot \frac{1}{2}(1+x)^{-1/2} + (1+x)^{1/2}$ | B1 | $x \cdot \frac{1}{2}(1+x)^{-1/2}$ |
| $= \frac{1}{2}(1+x)^{-1/2}[x + 2(1+x)]$ | B1, M1 | $\ldots + (1+x)^{1/2}$; taking out common factor or common denominator |
| $= \dfrac{2+3x}{2\sqrt{1+x}}$ * | E1 [4] | www |
**Alternate method using $u = x+1$:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = x+1 \Rightarrow du/dx = 1$, $y = (u-1)u^{1/2} = u^{3/2} - u^{1/2}$ | M1 | |
| $\frac{dy}{du} = \frac{3}{2}u^{\frac{1}{2}} - \frac{1}{2}u^{-\frac{1}{2}}$ | A1 | |
| $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{3}{2}(x+1)^{\frac{1}{2}} - \frac{1}{2}(x+1)^{-\frac{1}{2}}$ | M1 | |
| $= \frac{1}{2}(x+1)^{-\frac{1}{2}}(3x+3-1)$ | M1 | taking out common factor or common denominator |
| $= \dfrac{2+3x}{2\sqrt{1+x}}$ * | E1 [4] | |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $dy/dx = 0$ when $3x + 2 = 0$ | M1, A1cao | |
| $x = -\frac{2}{3}$ | | o. |
| $y = -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}$ | A1 | not $x \geq -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}$ (ft their $y$ value, even if approximate) |
| Range is $y \geq -\dfrac{2}{3}\sqrt{\dfrac{1}{3}}$ | B1 ft [4] | |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_{-1}^{0} x\sqrt{1+x}\, dx$ | | |
| Let $u = 1+x$, $du/dx = 1 \Rightarrow du = dx$ | M1 | $du = dx$ or $du/dx = 1$ or $dx/du = 1$ |
| When $x=-1$, $u=0$; when $x=0$, $u=1$ | B1 | changing limits – allow with no working shown provided limits are present and consistent with $dx$ and $du$ |
| $= \int_{0}^{1}(u-1)\sqrt{u}\, du$ | M1 | $(u-1)\sqrt{u}$ |
| $= \int_{0}^{1}(u^{3/2} - u^{1/2})\, du$ * | E1 | www – condone only final brackets missing, otherwise notation must be correct |
| $= \left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_{0}^{1}$ | B1 B1 | $\frac{2}{5}u^{5/2}$, $-\frac{2}{3}u^{3/2}$ (oe) |
| substituting correct limits | M1 | (can imply the zero limit) |
| $= \pm\dfrac{4}{15}$ | A1cao [8] | $\pm\frac{4}{15}$ or $\pm 0.27$ or better, not $0.26$ |
---2 Fig. 7 shows part of the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = x \sqrt { 1 + x }$. The curve meets the $x$-axis at the origin and at the point P .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-2_487_875_487_624}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Verify that the point P has coordinates $( - 1,0 )$. Hence state the domain of the function $\mathrm { f } ( x )$.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }$.\\
(iii) Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.\\
(iv) Use the substitution $u = 1 + x$ to show that
$$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l }
u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } }
\end{array} \right) \mathrm { d } u .$$
Hence find the area of the region enclosed by the curve and the $x$-axis.
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}