| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 11 |
| 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 two-part integration question requiring differentiation to find a tangent equation (routine verification) and then calculating an area between a curve and line using definite integration. While it involves fractional powers and requires careful setup of the integral bounds, these are well-practiced techniques at C1/C2 level with no novel problem-solving required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \dfrac{3}{4}x^2 - 4\sqrt{x} + 7 \Rightarrow \dfrac{dy}{dx} = \dfrac{3}{2}x - 2x^{-0.5}\) | M1A1 | M1: Differentiates to obtain at least one correct power for one of the terms in \(x\) (may be un-simplified), e.g. \(x^2 \to x^{2-1}\) or \(\sqrt{x} \to x^{\frac{1}{2}-1}\). A1: Correct derivative, allow un-simplified e.g. \(2\times\dfrac{3}{4}x^{2-1}\) or \(-4\times\dfrac{1}{2}x^{\frac{1}{2}-1}\) |
| At \(x=4\): \(\dfrac{dy}{dx} = \dfrac{3}{2}(4) - 2(4)^{-0.5} = \ldots\) | M1 | Substitutes \(x=4\) into a changed function in an attempt to find the gradient |
| \(y - 11 = \text{"5"}(x-4)\) or \(y = mx+c \Rightarrow 11 = \text{"5"}\times4 + c \Rightarrow c = \ldots\) | dM1 | Correct straight line method using \((4,11)\) correctly placed and their \(dy/dx\) at \(x=4\) for the tangent not the normal. If using \(y=mx+c\), must reach as far as finding a value for \(c\). Dependent on the previous M. |
| \(y = 5x - 9\) | A1* | Correct printed equation with no errors seen. Beware of the "5" appearing from wrong working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int \dfrac{3}{4}x^2 - 4\sqrt{x} + 7\, dx = \dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} + 7x\ (+c)\) | M1A1 | M1: \(x^n \to x^{n+1}\) on any term, may be un-simplified e.g. \(x^2\to x^{2+1}\), \(x^{0.5}\to x^{0.5+1}\), \(7\to 7x^1\). A1: Correct integration, may be un-simplified e.g. \(\frac{1}{3}\times\frac{3}{4}x^{2+1}\), \(-\frac{2}{3}\times4x^{0.5+1}\), \(7x^1\) and \(+c\) is not required. |
| Tangent meets \(x\) axis at \(x = 1.8\) | B1 | This may be embedded within a triangle area below or may be seen on a diagram |
| Area of triangle \(= \dfrac{1}{2}\times(4 - \text{'1.8'})\times11 = (12.1)\) — correct method for area of triangle, look for \(\dfrac{1}{2}\times(4-\text{'1.8'})\times11\); this may be implied by the evaluation of \(\displaystyle\int_{\text{'1.8'}}^{4} 5x-9\, dx = \left[5\dfrac{x^2}{2} - 9x\right]_{\text{'1.8'}}^{4}\) | M1 | Correct method for the area of a triangle |
| \(\left(\dfrac{1}{4}4^3 - \dfrac{8}{3}\times4^{1.5} + 7\times4\right) - \left(\dfrac{1}{4}1^3 - \dfrac{8}{3}\times1^{1.5} + 7\times1\right) - \text{'12.1'}\) | ddM1 | Correct method for area \(= A + B + C -\) Area \(C\). Dependent on both previous method marks. |
| \(= \text{awrt } 5.98\) | A1 | Area of \(R =\) awrt \(5.98\) or allow the exact answer of \(\dfrac{359}{60}\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int \dfrac{3}{4}x^2 - 4\sqrt{x} + 7\, dx = \dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} + 7x\ (+c)\) | M1A1 | M1: \(x^n \to x^{n+1}\) on any term, may be un-simplified. A1: Correct integration, may be un-simplified; \(+c\) not required. |
| Tangent meets \(x\) axis at \(x=1.8\) | B1 | This may be seen on a diagram |
| \(\pm\displaystyle\int_{\text{'1.8'}}^{4}\left(\dfrac{3}{4}x^2 - 4\sqrt{x}+7\right)-(5x-9)\,dx = \pm\left[\dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} - \dfrac{5x^2}{2}+16x\right]_{\text{'1.8'}}^{4} = \dfrac{56}{3}-15.7182\ldots\ (= 2.9485\ldots)\) | M1 | Attempts to integrate "curve \(-\) line" or "line \(-\) curve", substitute limits "1.8" and 4 and subtracts |
| \(\left(\left(\dfrac{1}{4}\text{"1.8"}^3 - \dfrac{8}{3}\text{"1.8"}^{1.5}+7\times\text{"1.8"}\right)-\left(\dfrac{1}{4}1^3 - \dfrac{8}{3}1^{1.5}+7\times1\right)+\text{'2.9485...'}\right)\) | ddM1 | Correct method for area \(= A + B\). Dependent on both previous method marks. |
| \(= \text{awrt } 5.98\) | A1 | Area of \(R =\) awrt \(5.98\) or allow exact answer of \(\dfrac{359}{60}\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm\left(\dfrac{3}{4}x^2 - 4\sqrt{x}+7-5x+9\right) = \pm\left(\dfrac{3}{4}x^2-4\sqrt{x}-5x+16\right)\) then \(\pm\displaystyle\int\dfrac{3}{4}x^2-4\sqrt{x}-5x+16\,dx = \pm\left(\dfrac{1}{4}x^3-\dfrac{8}{3}x^{1.5}-\dfrac{5x^2}{2}\right)+kx\ (+c)\) | M1A1 | M1: \(x^n\to x^{n+1}\) on any term, may be un-simplified. A1: Correct integration as shown, may be un-simplified for coefficients and powers; \(+c\) not required. |
| Tangent meets \(x\) axis at \(x=1.8\) | B1 | This may be embedded within a triangle area below or may be seen on a diagram |
| Area of triangle \(= \dfrac{1}{2}\times(\text{'1.8'}-1)\times | 5\times1-9 | = (1.6)\) — correct method, look for \(\dfrac{1}{2}\times(\text{'1.8'}-1)\times |
| \(\left(\left(\dfrac{1}{4}4^3-\dfrac{8}{3}4^{1.5}-\dfrac{5\times4^2}{2}+16\times4\right)-\left(\dfrac{1}{4}1^3-\dfrac{8}{3}1^{1.5}-\dfrac{5\times1^2}{2}+16\times1\right)-\text{'1.6'}\right)\) | ddM1 | Correct method for area \(= A+B+D-\) Area \(D\). Dependent on both previous method marks. |
| \(= \text{awrt }5.98\) | A1 | Area of \(R =\) awrt \(5.98\) or allow exact answer of \(\dfrac{359}{60}\) or equivalent |
## Question 12(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \dfrac{3}{4}x^2 - 4\sqrt{x} + 7 \Rightarrow \dfrac{dy}{dx} = \dfrac{3}{2}x - 2x^{-0.5}$ | M1A1 | M1: Differentiates to obtain at least one correct power for one of the terms in $x$ (may be un-simplified), e.g. $x^2 \to x^{2-1}$ or $\sqrt{x} \to x^{\frac{1}{2}-1}$. A1: Correct derivative, allow un-simplified e.g. $2\times\dfrac{3}{4}x^{2-1}$ or $-4\times\dfrac{1}{2}x^{\frac{1}{2}-1}$ |
| At $x=4$: $\dfrac{dy}{dx} = \dfrac{3}{2}(4) - 2(4)^{-0.5} = \ldots$ | M1 | Substitutes $x=4$ into a changed function in an attempt to find the gradient |
| $y - 11 = \text{"5"}(x-4)$ or $y = mx+c \Rightarrow 11 = \text{"5"}\times4 + c \Rightarrow c = \ldots$ | dM1 | Correct straight line method using $(4,11)$ correctly placed and their $dy/dx$ at $x=4$ for the tangent **not the normal**. If using $y=mx+c$, must reach as far as finding a value for $c$. **Dependent on the previous M.** |
| $y = 5x - 9$ | A1* | Correct printed equation with no errors seen. **Beware of the "5" appearing from wrong working.** |
> **Important Note:** Following a correct derivative, if candidate states $x=4$ so $dy/dx = 5$, this is fine if they then complete correctly — allow full marks. However, following a correct derivative, if the candidate **just** states $dy/dx = 5$ and then proceeds to obtain the correct straight line equation, the final mark can be withheld. Some evidence is needed that the candidate is considering the gradient at $x=4$.
---
## Question 12(b) — Way 1
*(Finds area under curve between 1 and 4 and subtracts triangle C)*
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int \dfrac{3}{4}x^2 - 4\sqrt{x} + 7\, dx = \dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} + 7x\ (+c)$ | M1A1 | M1: $x^n \to x^{n+1}$ on any term, may be un-simplified e.g. $x^2\to x^{2+1}$, $x^{0.5}\to x^{0.5+1}$, $7\to 7x^1$. A1: Correct integration, may be un-simplified e.g. $\frac{1}{3}\times\frac{3}{4}x^{2+1}$, $-\frac{2}{3}\times4x^{0.5+1}$, $7x^1$ and $+c$ is not required. |
| Tangent meets $x$ axis at $x = 1.8$ | B1 | This may be embedded within a triangle area below or may be seen on a diagram |
| Area of triangle $= \dfrac{1}{2}\times(4 - \text{'1.8'})\times11 = (12.1)$ — correct method for area of triangle, look for $\dfrac{1}{2}\times(4-\text{'1.8'})\times11$; this may be implied by the evaluation of $\displaystyle\int_{\text{'1.8'}}^{4} 5x-9\, dx = \left[5\dfrac{x^2}{2} - 9x\right]_{\text{'1.8'}}^{4}$ | M1 | Correct method for the area of a triangle |
| $\left(\dfrac{1}{4}4^3 - \dfrac{8}{3}\times4^{1.5} + 7\times4\right) - \left(\dfrac{1}{4}1^3 - \dfrac{8}{3}\times1^{1.5} + 7\times1\right) - \text{'12.1'}$ | ddM1 | Correct method for area $= A + B + C -$ Area $C$. **Dependent on both previous method marks.** |
| $= \text{awrt } 5.98$ | A1 | Area of $R =$ awrt $5.98$ or allow the exact answer of $\dfrac{359}{60}$ or equivalent |
---
## Question 12(b) — Way 2
*(Finds area under curve between 1 and "1.8" and adds "line − curve" or "curve − line" between "1.8" and 4)*
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int \dfrac{3}{4}x^2 - 4\sqrt{x} + 7\, dx = \dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} + 7x\ (+c)$ | M1A1 | M1: $x^n \to x^{n+1}$ on any term, may be un-simplified. A1: Correct integration, may be un-simplified; $+c$ not required. |
| Tangent meets $x$ axis at $x=1.8$ | B1 | This may be seen on a diagram |
| $\pm\displaystyle\int_{\text{'1.8'}}^{4}\left(\dfrac{3}{4}x^2 - 4\sqrt{x}+7\right)-(5x-9)\,dx = \pm\left[\dfrac{1}{4}x^3 - \dfrac{8}{3}x^{1.5} - \dfrac{5x^2}{2}+16x\right]_{\text{'1.8'}}^{4} = \dfrac{56}{3}-15.7182\ldots\ (= 2.9485\ldots)$ | M1 | Attempts to integrate "curve $-$ line" or "line $-$ curve", substitute limits "1.8" and 4 and subtracts |
| $\left(\left(\dfrac{1}{4}\text{"1.8"}^3 - \dfrac{8}{3}\text{"1.8"}^{1.5}+7\times\text{"1.8"}\right)-\left(\dfrac{1}{4}1^3 - \dfrac{8}{3}1^{1.5}+7\times1\right)+\text{'2.9485...'}\right)$ | ddM1 | Correct method for area $= A + B$. **Dependent on both previous method marks.** |
| $= \text{awrt } 5.98$ | A1 | Area of $R =$ awrt $5.98$ or allow exact answer of $\dfrac{359}{60}$ or equivalent |
---
## Question 12(b) — Way 3
*(Uses "line − curve" or "curve − line" between 1 and 4 and subtracts triangle below $x$ axis)*
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\left(\dfrac{3}{4}x^2 - 4\sqrt{x}+7-5x+9\right) = \pm\left(\dfrac{3}{4}x^2-4\sqrt{x}-5x+16\right)$ then $\pm\displaystyle\int\dfrac{3}{4}x^2-4\sqrt{x}-5x+16\,dx = \pm\left(\dfrac{1}{4}x^3-\dfrac{8}{3}x^{1.5}-\dfrac{5x^2}{2}\right)+kx\ (+c)$ | M1A1 | M1: $x^n\to x^{n+1}$ on any term, may be un-simplified. A1: Correct integration as shown, may be un-simplified for coefficients and powers; $+c$ not required. |
| Tangent meets $x$ axis at $x=1.8$ | B1 | This may be embedded within a triangle area below or may be seen on a diagram |
| Area of triangle $= \dfrac{1}{2}\times(\text{'1.8'}-1)\times|5\times1-9| = (1.6)$ — correct method, look for $\dfrac{1}{2}\times(\text{'1.8'}-1)\times|5\times1-9|$ | M1 | Correct method for the area of a triangle |
| $\left(\left(\dfrac{1}{4}4^3-\dfrac{8}{3}4^{1.5}-\dfrac{5\times4^2}{2}+16\times4\right)-\left(\dfrac{1}{4}1^3-\dfrac{8}{3}1^{1.5}-\dfrac{5\times1^2}{2}+16\times1\right)-\text{'1.6'}\right)$ | ddM1 | Correct method for area $= A+B+D-$ Area $D$. **Dependent on both previous method marks.** |
| $= \text{awrt }5.98$ | A1 | Area of $R =$ awrt $5.98$ or allow exact answer of $\dfrac{359}{60}$ or equivalent |12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f39ade34-32e2-4b5c-b80a-9663c6a65c87-20_775_1015_260_459}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 3 } { 4 } x ^ { 2 } - 4 \sqrt { x } + 7 , \quad x > 0$$
The point $P$ lies on $C$ and has coordinates $( 4,11 )$.\\
Line $l$ is the tangent to $C$ at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that $l$ has equation $y = 5 x - 9$
The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the line $x = 1$, the $x$-axis and the line $l$.
\item Find, by using calculus, the area of $R$, giving your answer to 2 decimal places.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q12 [11]}}