| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring finding area under a curve, then using the tangent equation to find the area of a triangle. The integration is straightforward (power rule), and the 'hence' part involves basic geometric reasoning (triangle area minus curve area). Slightly above average due to the multi-step nature and need to find the tangent equation, but all techniques are routine for C2. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_1^4 \left(x^{\frac{3}{2}}-1\right)dx = \left[\frac{2}{5}x^{\frac{5}{2}}-x\right]_1^4\) | M1 | Attempt integration. Increase in power by 1 for at least one term — allow the \(-1\) to disappear |
| Correct unsimplified integral | A1 | Coeff could be unsimplified e.g. \(\frac{1}{2.5}\). Could have \(+c\) present |
| \(= (12.8 - 4) - (0.4 - 1)\) | M1 | Attempt correct use of limits. Must be explicitly attempting \(F(4) - F(1)\), either by clear substitution of 4 and 1 or by showing at least \((8.8) - (-0.6)\). Allow M1 if \(+c\) still present in both \(F(4)\) and \(F(1)\), but M0 if \(c\) is now numerical |
| \(= 9\frac{2}{5}\) AG | A1 [4] | AG, so check method carefully. Allow \(\frac{47}{5}\) or 9.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m = \frac{3}{2} \times \sqrt{4} = 3\) | M1* | Attempt to find gradient at \((4, 7)\) using differentiation. Must be reasonable attempt at differentiation i.e. decrease the power by 1. Need to actually evaluate derivative at \(x = 4\) |
| \(y = 3x - 5\) | M1d* | Attempt to find point of intersection of tangent with \(x\)-axis or attempt to find base of triangle. Could attempt equation of tangent and use \(y = 0\). Could use equiv method with gradient e.g. \(3 = \frac{7}{4-x}\). Could just find base of triangle using gradient e.g. \(3 = \frac{7}{b}\) |
| Tangent crosses \(x\)-axis at \(\left(\frac{5}{3}, 0\right)\) | A1 | Obtain \(x = \frac{5}{3}\) as point of intersection or obtain \(\frac{7}{3}\) as base of triangle. Allow decimal equiv, such as 1.7, 1.67 or even 1.6 www. Allow M1M1A1 for \(x = \frac{5}{3}\) with no method shown |
| Area of triangle \(= \frac{1}{2} \times \left(4 - \frac{5}{3}\right) \times 7 = 8\frac{1}{6}\) | M1d** | Attempt complete method to find shaded area. Dependent on both previous M marks. Find area of triangle and subtract from \(9\frac{2}{5}\). Must have \(1 <\) their \(x < 4\), and area of triangle \(< 9\frac{2}{5}\). If using \(\int(3x-5)dx\) then limits must be 4 and their \(x\). M1 for area of trapezium — area between curve and \(y\)-axis |
| Shaded area \(= 9\frac{2}{5} - 8\frac{1}{6} = 1\frac{7}{30}\) | A1 [5] | Obtain \(1\frac{7}{30}\), or exact equiv. A0 for decimal answer (1.23), unless clearly a recurring decimal (but not e.g. 1.2333...) |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^4 \left(x^{\frac{3}{2}}-1\right)dx = \left[\frac{2}{5}x^{\frac{5}{2}}-x\right]_1^4$ | M1 | Attempt integration. Increase in power by 1 for at least one term — allow the $-1$ to disappear |
| Correct unsimplified integral | A1 | Coeff could be unsimplified e.g. $\frac{1}{2.5}$. Could have $+c$ present |
| $= (12.8 - 4) - (0.4 - 1)$ | M1 | Attempt correct use of limits. Must be explicitly attempting $F(4) - F(1)$, either by clear substitution of 4 and 1 or by showing at least $(8.8) - (-0.6)$. Allow M1 if $+c$ still present in both $F(4)$ and $F(1)$, but M0 if $c$ is now numerical |
| $= 9\frac{2}{5}$ **AG** | A1 [4] | **AG**, so check method carefully. Allow $\frac{47}{5}$ or 9.4 |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m = \frac{3}{2} \times \sqrt{4} = 3$ | M1* | Attempt to find gradient at $(4, 7)$ using differentiation. Must be reasonable attempt at differentiation i.e. decrease the power by 1. Need to actually evaluate derivative at $x = 4$ |
| $y = 3x - 5$ | M1d* | Attempt to find point of intersection of tangent with $x$-axis or attempt to find base of triangle. Could attempt equation of tangent and use $y = 0$. Could use equiv method with gradient e.g. $3 = \frac{7}{4-x}$. Could just find base of triangle using gradient e.g. $3 = \frac{7}{b}$ |
| Tangent crosses $x$-axis at $\left(\frac{5}{3}, 0\right)$ | A1 | Obtain $x = \frac{5}{3}$ as point of intersection or obtain $\frac{7}{3}$ as base of triangle. Allow decimal equiv, such as 1.7, 1.67 or even 1.6 www. Allow M1M1A1 for $x = \frac{5}{3}$ with no method shown |
| Area of triangle $= \frac{1}{2} \times \left(4 - \frac{5}{3}\right) \times 7 = 8\frac{1}{6}$ | M1d** | Attempt complete method to find shaded area. Dependent on both previous M marks. Find area of triangle and subtract from $9\frac{2}{5}$. Must have $1 <$ their $x < 4$, and area of triangle $< 9\frac{2}{5}$. If using $\int(3x-5)dx$ then limits must be 4 and their $x$. M1 for area of trapezium — area between curve and $y$-axis |
| Shaded area $= 9\frac{2}{5} - 8\frac{1}{6} = 1\frac{7}{30}$ | A1 [5] | Obtain $1\frac{7}{30}$, or exact equiv. A0 for decimal answer (1.23), unless clearly a recurring decimal (but not e.g. 1.2333...) |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{b2c1188d-881e-4fb5-bece-5a51006543c7-3_519_611_1087_712}
The diagram shows the curve $y = x ^ { \frac { 3 } { 2 } } - 1$, which crosses the $x$-axis at $( 1,0 )$, and the tangent to the curve at the point $( 4,7 )$.\\
(i) Show that $\int _ { 1 } ^ { 4 } \left( x ^ { \frac { 3 } { 2 } } - 1 \right) \mathrm { d } x = 9 \frac { 2 } { 5 }$.\\
(ii) Hence find the exact area of the shaded region enclosed by the curve, the tangent and the $x$-axis.
\hfill \mbox{\textit{OCR C2 2013 Q7 [9]}}