OCR C2 — Question 8 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeImproper integral evaluation
DifficultyStandard +0.3 Part (i) is a straightforward definite integral requiring polynomial integration and solving for k. Part (ii) is an improper integral but uses a standard technique (limit as upper bound approaches infinity) with a simple power function. Both parts are routine C2-level exercises requiring only direct application of standard methods, making this slightly easier than average.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits4.08c Improper integrals: infinite limits or discontinuous integrands

  1. (i) Given that
$$\int _ { 1 } ^ { 3 } \left( x ^ { 2 } - 2 x + k \right) d x = 8 \frac { 2 } { 3 }$$ find the value of the constant \(k\).
(ii) Evaluate $$\int _ { 2 } ^ { \infty } \frac { 6 } { x ^ { \frac { 5 } { 2 } } } \mathrm {~d} x$$ giving your answer in its simplest form.

Question 8:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Notes
\(\int_1^3 (x^2 - 2x + k)\ dx = \left[\frac{1}{3}x^3 - x^2 + kx\right]_1^3\)M1 A2
\(= (9 - 9 + 3k) - (\frac{1}{3} - 1 + k)\)M1
\(= 2k + \frac{2}{3}\)
\(\therefore 2k + \frac{2}{3} = 8\frac{2}{3}\)
\(k = 4\)M1 A1
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Notes
\(= \lim_{k \to \infty} \left[-4x^{-\frac{3}{2}}\right]_2^k\)M2 A1
\(= \lim_{k \to \infty} \left\{-\frac{4}{k^{\frac{3}{2}}} - \left(-\frac{4}{2\sqrt{2}}\right)\right\}\)M1
\(= \lim_{k \to \infty} \left(\sqrt{2} - \frac{4}{k^{\frac{3}{2}}}\right) = \sqrt{2}\)A1 (11)
# Question 8:

## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\int_1^3 (x^2 - 2x + k)\ dx = \left[\frac{1}{3}x^3 - x^2 + kx\right]_1^3$ | M1 A2 | |
| $= (9 - 9 + 3k) - (\frac{1}{3} - 1 + k)$ | M1 | |
| $= 2k + \frac{2}{3}$ | | |
| $\therefore 2k + \frac{2}{3} = 8\frac{2}{3}$ | | |
| $k = 4$ | M1 A1 | |

## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $= \lim_{k \to \infty} \left[-4x^{-\frac{3}{2}}\right]_2^k$ | M2 A1 | |
| $= \lim_{k \to \infty} \left\{-\frac{4}{k^{\frac{3}{2}}} - \left(-\frac{4}{2\sqrt{2}}\right)\right\}$ | M1 | |
| $= \lim_{k \to \infty} \left(\sqrt{2} - \frac{4}{k^{\frac{3}{2}}}\right) = \sqrt{2}$ | A1 | **(11)** |

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\begin{enumerate}
  \item (i) Given that
\end{enumerate}

$$\int _ { 1 } ^ { 3 } \left( x ^ { 2 } - 2 x + k \right) d x = 8 \frac { 2 } { 3 }$$

find the value of the constant $k$.\\
(ii) Evaluate

$$\int _ { 2 } ^ { \infty } \frac { 6 } { x ^ { \frac { 5 } { 2 } } } \mathrm {~d} x$$

giving your answer in its simplest form.\\

\hfill \mbox{\textit{OCR C2  Q8 [11]}}