OCR MEI Paper 1 Specimen — Question 8 7 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks7
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Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeDouble integration by parts
DifficultyStandard +0.3 This is a standard double integration by parts question requiring two applications of the technique with a polynomial and exponential. While it requires careful execution and algebraic manipulation, it follows a well-practiced routine with no conceptual surprises, making it slightly easier than average for A-level Further Maths students who have been trained on this method.
Spec1.08i Integration by parts
8 Find \(\int x ^ { 2 } \mathrm { e } ^ { 2 x } \mathrm {~d} x\).
Question 8:
AnswerMarks Guidance
Let \(u = x^2,\ u' = 2x,\ v' = e^{2x},\ v = \frac{1}{2}e^{2x}\)M1A1
\(\int x^2e^{2x}dx = \frac{1}{2}x^2e^{2x} - \int 2x \cdot \frac{1}{2}e^{2x}dx = \frac{1}{2}x^2e^{2x} - \int xe^{2x}dx\)A1
Let \(u = x,\ u' = 1,\ v' = e^{2x},\ v = \frac{1}{2}e^{2x}\)M1
\(\int xe^{2x}dx = \frac{1}{2}xe^{2x} - \int\frac{1}{2}e^{2x}dx\)A1
\(= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}(+c)\)A1
so \(\int x^2e^{2x}dx = \frac{1}{2}x^2e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + c\)A1 [7] Do not award if no \(+c\) 
## Question 8:
Let $u = x^2,\ u' = 2x,\ v' = e^{2x},\ v = \frac{1}{2}e^{2x}$ | **M1A1** |
$\int x^2e^{2x}dx = \frac{1}{2}x^2e^{2x} - \int 2x \cdot \frac{1}{2}e^{2x}dx = \frac{1}{2}x^2e^{2x} - \int xe^{2x}dx$ | **A1** |
Let $u = x,\ u' = 1,\ v' = e^{2x},\ v = \frac{1}{2}e^{2x}$ | **M1** |
$\int xe^{2x}dx = \frac{1}{2}xe^{2x} - \int\frac{1}{2}e^{2x}dx$ | **A1** |
$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}(+c)$ | **A1** |
so $\int x^2e^{2x}dx = \frac{1}{2}x^2e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + c$ | **A1** [7] | Do not award if no $+c$

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8 Find $\int x ^ { 2 } \mathrm { e } ^ { 2 x } \mathrm {~d} x$.

\hfill \mbox{\textit{OCR MEI Paper 1  Q8 [7]}}
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