OCR MEI C3 — Question 1 5 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeIntegration of x^n·ln(x)
DifficultyStandard +0.3 This is a straightforward application of integration by parts with a standard x^n·ln(x) form. The technique is routine for C3 level: let u=ln(x), dv=x³dx, then integrate and evaluate limits. Slightly above average difficulty only because it requires careful execution and exact value calculation, but follows a well-practiced pattern.
Spec1.08i Integration by parts
1 Find the exact value of \(\int ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x\).
Question 1:
M1: let \(u = \ln x\), \(\frac{dv}{dx} = x^3\), \(\frac{du}{dx} = \frac{1}{x}\), \(v = \frac{1}{4}x^4\)
A1: \(\int_1^2 x^3\ln x \, dx = \left[\frac{1}{4}x^4\ln x\right]_1^2 - \int_1^2 \frac{1}{4}x^4 \cdot \frac{1}{x} \, dx\)
M1 dep: \(\left[\frac{1}{4}x^4\ln x - \int_1^2 \frac{1}{4}x^3 \, dx\right]_1^2\)
A1 cao: \(\left[\frac{1}{4}x^4\ln x - \frac{1}{16}x^4\right]_1^2\)
A1 cao: \(= 4\ln 2 - \frac{15}{16}\)
Guidance notes:
\(u\), \(\dot{u}\), \(v\), \(\dot{v}\) all correct
\(\frac{1}{4}x^4\ln x - \int \frac{1}{4}x^4 \cdot \frac{1}{x} \, [dx]\)
Simplifying \(\frac{x^4}{x} = x^3\) in second term (soi)
\(\frac{1}{4}x^4\ln x - \frac{1}{16}x^4\) o.e.
o.e. must be exact, but can isw (ignore limits)
Dep on 1st M1
Must evaluate \(\ln 1 = 0\) and combine
\(-1 + \frac{1}{16}\)
Question 1:

M1: let $u = \ln x$, $\frac{dv}{dx} = x^3$, $\frac{du}{dx} = \frac{1}{x}$, $v = \frac{1}{4}x^4$

A1: $\int_1^2 x^3\ln x \, dx = \left[\frac{1}{4}x^4\ln x\right]_1^2 - \int_1^2 \frac{1}{4}x^4 \cdot \frac{1}{x} \, dx$

M1 dep: $\left[\frac{1}{4}x^4\ln x - \int_1^2 \frac{1}{4}x^3 \, dx\right]_1^2$

A1 cao: $\left[\frac{1}{4}x^4\ln x - \frac{1}{16}x^4\right]_1^2$

A1 cao: $= 4\ln 2 - \frac{15}{16}$

Guidance notes:

$u$, $\dot{u}$, $v$, $\dot{v}$ all correct

$\frac{1}{4}x^4\ln x - \int \frac{1}{4}x^4 \cdot \frac{1}{x} \, [dx]$

Simplifying $\frac{x^4}{x} = x^3$ in second term (soi)

$\frac{1}{4}x^4\ln x - \frac{1}{16}x^4$ o.e.

o.e. must be exact, but can isw (ignore limits)

Dep on 1st M1

Must evaluate $\ln 1 = 0$ and combine

$-1 + \frac{1}{16}$
1 Find the exact value of $\int ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x$.

\hfill \mbox{\textit{OCR MEI C3  Q1 [5]}}
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Q1 5 Q2 18 Q3 5 Q4 5 Q5 18