| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic numerical integration skills. Part (a) requires simple substitution into a square root function, part (b) is a standard trapezium rule application with values provided, and parts (c)(i)-(ii) involve elementary manipulations of the integral using constant factors and linearity. All steps are routine procedural work with no problem-solving or conceptual challenges required. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| \(x\) | - 2 | 0 | 2 | 4 | 6 |
| \(y\) | 0 | 1.4142 | 2 | 2.8284 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\approx 2.4495\) | B1 | Awrt \(2.4495\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Strip width \(= 2\) | B1 | May be implied by sight of \(\frac{2}{2}\{\ldots\}\) or \(1\times\{\ldots\}\) |
| Area \(\approx \frac{2}{2}\{0 + 2.8284 + 2\times(1.4142 + 2 + 2.4495)\} \approx 14.556\) | M1A1 | M1: correct form of bracket within trapezium rule; A1: awrt \(14.556\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}(b) = 7.278\) | B1ft | Follow through on answer to (b); allow awrt \(7.28\); also allow from adapting trapezium rule with all terms halved |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_{-2}^{6} 2\, dx = [2x]_{-2}^{6} = 12-(-4) = 16\) | M1 | Attempt to integrate \(2\); may be embedded in longer integral; alternatively area of rectangle height 2, length \((6-(-2))\); do not allow \(\int 2\,dx = 2\) |
| \(\int_{-2}^{6}(2+\sqrt{x+2})\,dx = 16 + (b) = 30.556\) | dM1 A1ft | dM1 (dep): adds integral of 2 to answer for (b); A1ft: \(16+(b)\); allow awrt \(30.56\); accept for all 3 marks if no incorrect working seen |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\approx 2.4495$ | B1 | Awrt $2.4495$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Strip width $= 2$ | B1 | May be implied by sight of $\frac{2}{2}\{\ldots\}$ or $1\times\{\ldots\}$ |
| Area $\approx \frac{2}{2}\{0 + 2.8284 + 2\times(1.4142 + 2 + 2.4495)\} \approx 14.556$ | M1A1 | M1: correct form of bracket within trapezium rule; A1: awrt $14.556$ |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}(b) = 7.278$ | B1ft | Follow through on answer to (b); allow awrt $7.28$; also allow from adapting trapezium rule with all terms halved |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{-2}^{6} 2\, dx = [2x]_{-2}^{6} = 12-(-4) = 16$ | M1 | Attempt to integrate $2$; may be embedded in longer integral; alternatively area of rectangle height 2, length $(6-(-2))$; do not allow $\int 2\,dx = 2$ |
| $\int_{-2}^{6}(2+\sqrt{x+2})\,dx = 16 + (b) = 30.556$ | dM1 A1ft | dM1 (dep): adds integral of 2 to answer for (b); A1ft: $16+(b)$; allow awrt $30.56$; accept for all 3 marks if no incorrect working seen |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-05_476_1338_251_360}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \sqrt { x + 2 } , x \geqslant - 2$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the line $x = 6$
The table below shows corresponding values of $x$ and $y$ for $y = \sqrt { x + 2 }$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 2 & 0 & 2 & 4 & 6 \\
\hline
$y$ & 0 & 1.4142 & 2 & & 2.8284 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above, giving the missing value of $y$ to 4 decimal places.
\item Use the trapezium rule, with all of the values of $y$ in the completed table, to find an approximate value for the area of $R$, giving your answer to 3 decimal places.
Use your answer to part (b) to find approximate values of
\item \begin{enumerate}[label=(\roman*)]
\item $\int _ { - 2 } ^ { 6 } \frac { \sqrt { x + 2 } } { 2 } \mathrm {~d} x$
\item $\int _ { - 2 } ^ { 6 } ( 2 + \sqrt { x + 2 } ) \mathrm { d } x$
\begin{center}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q4 [8]}}