OCR MEI Paper 3 2021 November — Question 7 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2021
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.3 This is a straightforward application of integration by parts with standard functions (polynomial times trigonometric). It requires only one application of the formula with u=x and dv=cos(2x)dx, making it slightly easier than average but still requiring proper technique and care with the chain rule when integrating cos(2x).
Spec1.08i Integration by parts
7 Determine \(\int x \cos 2 x \mathrm {~d} x\).
Question 7:
AnswerMarks Guidance
Let \(u = x\), \(\frac{dv}{dx} = \cos 2x\)M1 (3.1a) Parts with \(u = x\), \(\frac{dv}{dx} = \cos 2x\)
\(v = \frac{1}{2}\sin 2x\); \(\int x\cos 2x\, dx = \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x\, dx\)M1 (1.1) Allow if \(-\left[\frac{1}{2}x\sin 2x - \int\frac{1}{2}\sin 2x\, dx\right]\) or if 1 error
\(\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + c\)A1 (2.5) \(+c\) needed for A1
[3 marks]
## Question 7:

Let $u = x$, $\frac{dv}{dx} = \cos 2x$ | M1 (3.1a) | Parts with $u = x$, $\frac{dv}{dx} = \cos 2x$
$v = \frac{1}{2}\sin 2x$; $\int x\cos 2x\, dx = \frac{1}{2}x\sin 2x - \int \frac{1}{2}\sin 2x\, dx$ | M1 (1.1) | Allow if $-\left[\frac{1}{2}x\sin 2x - \int\frac{1}{2}\sin 2x\, dx\right]$ or if 1 error
$\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + c$ | A1 (2.5) | $+c$ needed for A1
[3 marks]

---
7 Determine $\int x \cos 2 x \mathrm {~d} x$.

\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q7 [3]}}
This paper (15 questions)
View full paper
Q1 5 Q2 2 Q3 7 Q4 3 Q5 8 Q6 4 Q7 3 Q8 3 Q9 11 Q10 9 Q11 5 Q12 3 Q13 3 Q14 5 Q15 4