OCR C2 2010 June — Question 6 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2010
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea under polynomial curve
DifficultyModerate -0.8 This is a straightforward C2 question testing basic integration skills across three standard parts: (a) direct polynomial integration with limits, (b) simple indefinite integration of a polynomial in y, and (c) an improper integral that reduces to routine calculation. All parts are textbook exercises requiring only recall and mechanical application of integration rules, making this easier than average for A-level.
Spec1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals

6
  1. Use integration to find the exact area of the region enclosed by the curve \(y = x ^ { 2 } + 4 x\), the \(x\)-axis and the lines \(x = 3\) and \(x = 5\).
  2. Find \(\int ( 2 - 6 \sqrt { y } ) \mathrm { d } y\).
  3. Evaluate \(\int _ { 1 } ^ { \infty } \frac { 8 } { x ^ { 3 } } \mathrm {~d} x\).

Question 6:
Part (a)
AnswerMarks Guidance
\(\int_3^5(x^2+4x)dx = \left[\frac{1}{3}x^3 + 2x^2\right]_3^5\)M1 Attempt integration
A1Obtain \(\frac{1}{3}x^3 + 2x^2\)
\(= \left(\frac{125}{3} + 50\right) - (9 + 18)\)M1 Use limits \(x = 3, 5\) – correct order & subtraction
\(= 64\frac{2}{3}\)A1 Obtain \(64\frac{2}{3}\) or any exact equivalent
Part (b)
AnswerMarks Guidance
\(\int(2 - 6\sqrt{y})dy = 2y - 4y^{\frac{3}{2}} + c\)B1 State \(2y\)
M1Obtain \(ky^{\frac{3}{2}}\)
A1Obtain \(-4y^{\frac{3}{2}}\) (condone absence of \(+c\))
Part (c)
AnswerMarks Guidance
\(\int_1^\infty 8x^{-3}dx = \left[\frac{-4}{x^2}\right]_1^\infty\)B1 State or imply \(\frac{1}{x^3} = x^{-3}\)
M1Attempt integration of \(kx^n\)
A1Obtain correct \(-4x^{-2}\) \((+c)\)
\(= (0) - (-4) = 4\)A1 ft Obtain 4 (or \(-k\) following their \(kx^{-2}\))
## Question 6:

### Part (a)
| $\int_3^5(x^2+4x)dx = \left[\frac{1}{3}x^3 + 2x^2\right]_3^5$ | M1 | Attempt integration |
| | A1 | Obtain $\frac{1}{3}x^3 + 2x^2$ |
| $= \left(\frac{125}{3} + 50\right) - (9 + 18)$ | M1 | Use limits $x = 3, 5$ – correct order & subtraction |
| $= 64\frac{2}{3}$ | A1 | Obtain $64\frac{2}{3}$ or any exact equivalent |

### Part (b)
| $\int(2 - 6\sqrt{y})dy = 2y - 4y^{\frac{3}{2}} + c$ | B1 | State $2y$ |
| | M1 | Obtain $ky^{\frac{3}{2}}$ |
| | A1 | Obtain $-4y^{\frac{3}{2}}$ (condone absence of $+c$) |

### Part (c)
| $\int_1^\infty 8x^{-3}dx = \left[\frac{-4}{x^2}\right]_1^\infty$ | B1 | State or imply $\frac{1}{x^3} = x^{-3}$ |
| | M1 | Attempt integration of $kx^n$ |
| | A1 | Obtain correct $-4x^{-2}$ $(+c)$ |
| $= (0) - (-4) = 4$ | A1 ft | Obtain 4 (or $-k$ following their $kx^{-2}$) |

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6
\begin{enumerate}[label=(\alph*)]
\item Use integration to find the exact area of the region enclosed by the curve $y = x ^ { 2 } + 4 x$, the $x$-axis and the lines $x = 3$ and $x = 5$.
\item Find $\int ( 2 - 6 \sqrt { y } ) \mathrm { d } y$.
\item Evaluate $\int _ { 1 } ^ { \infty } \frac { 8 } { x ^ { 3 } } \mathrm {~d} x$.
\end{enumerate}

\hfill \mbox{\textit{OCR C2 2010 Q6 [11]}}