OCR C2 — Question 6 9 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeFind curve from gradient
DifficultyModerate -0.8 This is a straightforward two-part question testing basic integration skills. Part (i) requires standard integration of polynomial terms and evaluation at limits. Part (ii) is a routine 'find the curve from gradient' problem requiring integration and using an initial condition to find the constant. Both parts involve only direct application of standard techniques with no problem-solving or insight required, making it easier than average.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits

6. (i) Evaluate $$\int _ { 2 } ^ { 4 } \left( 2 - \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x$$ (ii) Given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { 3 } + 1$$ and that \(y = 3\) when \(x = 0\), find the value of \(y\) when \(x = 2\).

Question 6:
Part (i):
AnswerMarks
\(= [2x + x^{-1}]_2^4\)M1 A1
\(= (8 + \frac{1}{4}) - (4 + \frac{1}{2}) = 3\frac{3}{4}\)M1 A1
Part (ii):
AnswerMarks Guidance
\(y = \int (2x^3 + 1)\, dx\)
\(y = \frac{1}{2}x^4 + x + c\)M1 A1
\(x = 0, y = 3 \therefore c = 3\)B1
\(y = \frac{1}{2}x^4 + x + 3\)
when \(x = 2\), \(y = 8 + 2 + 3 = 13\)M1 A1 (9)
# Question 6:

## Part (i):
$= [2x + x^{-1}]_2^4$ | M1 A1 |
$= (8 + \frac{1}{4}) - (4 + \frac{1}{2}) = 3\frac{3}{4}$ | M1 A1 |

## Part (ii):
$y = \int (2x^3 + 1)\, dx$ |  |
$y = \frac{1}{2}x^4 + x + c$ | M1 A1 |
$x = 0, y = 3 \therefore c = 3$ | B1 |
$y = \frac{1}{2}x^4 + x + 3$ |  |
when $x = 2$, $y = 8 + 2 + 3 = 13$ | M1 A1 | **(9)**

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6. (i) Evaluate

$$\int _ { 2 } ^ { 4 } \left( 2 - \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x$$

(ii) Given that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { 3 } + 1$$

and that $y = 3$ when $x = 0$, find the value of $y$ when $x = 2$.\\

\hfill \mbox{\textit{OCR C2  Q6 [9]}}