| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Complete table then estimate |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic numerical integration. Part (a) requires simple substitution into a calculator, part (b) is direct application of the trapezium rule formula with given values, and part (c) requires recognizing that adding a constant to an integrand adds (constant × width) to the integral result. All steps are routine with no problem-solving or conceptual challenges beyond standard C2 content. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.5 | 1 | 1.5 | 2 |
| \(y\) | 1 | 2.821 | 12.502 | 26.585 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\{\)At \(x=1,\}\) \(y=6\) (allow 6.000 or even 6.00) | B1 cao | |
| (b) \(\frac{1}{2}\times 0.5\); | B1 oe | |
| \(\{1+26.585+2(2.821+\text{their } 6+12.502)\}\) | M1 ∆1n | For structure of \(\{\ldots\}\) |
| \(\frac{1}{2}\times 0.5\left\{1+26.585+2(2.821+6+12.502)\right\}=\frac{1}{2}(70.231)=17.557\ldots=\text{awrt } 17.56\) | A1 | |
| (c) \(10+"17.56"="27.56"\) | B1 ft |
| (a) $\{$At $x=1,\}$ $y=6$ (allow 6.000 or even 6.00) | B1 cao | |
|---|---|---|
| (b) $\frac{1}{2}\times 0.5$; | B1 oe | |
| $\{1+26.585+2(2.821+\text{their } 6+12.502)\}$ | M1 ∆1n | For structure of $\{\ldots\}$ |
| $\frac{1}{2}\times 0.5\left\{1+26.585+2(2.821+6+12.502)\right\}=\frac{1}{2}(70.231)=17.557\ldots=\text{awrt } 17.56$ | A1 | |
| (c) $10+"17.56"="27.56"$ | B1 ft | |
**Notes:**
- (a) B1: 6
- (b) B1: for using $\frac{1}{2}\times 0.5$ or $\frac{1}{4}$ or equivalent; M1: requires correct $\{\ldots\}$ bracket structure. Needs first bracket to contain first $y$ value plus last $y$ value and second bracket to be multiplied by 2 and to be summation of remaining $y$ values in table with no additional values. If only mistake is copying error or is to omit one value from 2nd bracket this may be regarded as a slip and M mark can be allowed (An extra repeated term forfeits the M mark however); M0 if values used in brackets are $x$ values instead of $y$ values; A1 ft: for correct bracket $\{\ldots\}$ following through candidate's $y$ value found in part (a)
- (c) B1 ft: $10+$ their answer to part (b). (May be obtained by using trapezium rule again with all values for $y$ increased by 5)3. (a) $\quad y = 5 ^ { x } + \log _ { 2 } ( x + 1 ) , \quad 0 \leqslant x \leqslant 2$
Complete the table below, by giving the value of $y$ when $x = 1$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.5 & 1 & 1.5 & 2 \\
\hline
$y$ & 1 & 2.821 & & 12.502 & 26.585 \\
\hline
\end{tabular}
\end{center}
(b) Use the trapezium rule, with all the values of $y$ from the completed table, to find an approximate value for
$$\int _ { 0 } ^ { 2 } \left( 5 ^ { x } + \log _ { 2 } ( x + 1 ) \right) \mathrm { d } x$$
giving your answer to 2 decimal places.\\
(c) Use your answer to part (b) to find an approximate value for
$$\int _ { 0 } ^ { 2 } \left( 5 + 5 ^ { x } + \log _ { 2 } ( x + 1 ) \right) d x$$
giving your answer to 2 decimal places.
\hfill \mbox{\textit{Edexcel C2 2017 Q3 [6]}}