Pre-U Pre-U 9794/2 2011 June — Question 3 5 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2011
SessionJune
Marks5
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.8 This is a straightforward application of integration by parts with a simple polynomial and trigonometric function. It requires only one application of the technique with standard choices (u=x, dv=sin 3x dx), making it easier than average and essentially a textbook exercise testing basic recall of the method.
Spec1.08i Integration by parts
Use integration by parts to find \(\int x \sin 3x \, dx\). [5]
AnswerMarks Guidance
\(\frac{dv}{dx} = \sin 3x\), \(u = x\)M1
\(v = -\frac{1}{3}\cos 3x\), \(\frac{du}{dx} = 1\)A1
Obtain an expression of the form \(f(x) \pm \int g(x)dx\)M1
Obtain \(-\frac{x}{3}\cos 3x + \int \frac{1}{3}\cos 3x \, dx\)A1 ∨
\(= -\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + c\)CAO A1 
$\frac{dv}{dx} = \sin 3x$, $u = x$ | M1
$v = -\frac{1}{3}\cos 3x$, $\frac{du}{dx} = 1$ | A1
Obtain an expression of the form $f(x) \pm \int g(x)dx$ | M1
Obtain $-\frac{x}{3}\cos 3x + \int \frac{1}{3}\cos 3x \, dx$ | A1 ∨
$= -\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + c$ | CAO | A1 | [5]
Use integration by parts to find $\int x \sin 3x \, dx$. [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q3 [5]}}
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