Standard +0.3 This is a straightforward integration by parts question with a standard form (power of x times ln x). The substitution u = ln x, dv = x^(-1/3) dx is routine, and the integration limits are simple. While it requires careful algebraic manipulation and evaluation at x=1 and x=8, it follows a well-practiced template with no conceptual surprises, making it slightly easier than average.
4 Find the exact value of \(\int _ { 1 } ^ { 8 } \frac { 1 } { \sqrt [ 3 ] { x } } \ln x \mathrm {~d} x\), giving your answer in the form \(A \ln 2 + B\), where \(A\) and \(B\) are constants to be found.
\(Ax^{\frac{2}{3}}\ln x - \int Bx^{\frac{2}{3}} \times \frac{1}{x}\,dx\)
M1*
\(A\) and \(B\) are non-zero constants
\(\frac{3}{2}x^{\frac{2}{3}}\ln x - \int\frac{3}{2}x^{\frac{2}{3}} \times \frac{1}{x}\,dx\)
A1
ignore \(+c\); NB \(\frac{3}{2}x^{\frac{2}{3}}\ln x - \int\frac{3}{2}x^{-\frac{1}{3}}dx\); allow both marks if \(dx\) omitted
\(F[x] = \frac{3}{2}x^{\frac{2}{3}}\ln x - \frac{\frac{3}{2}}{\frac{2}{3}}x^{\frac{2}{3}}\)
A1
ignore limits for first three marks
\(F[8] - F[1]\)
M1*dep
dependent on integration of their \(\frac{3}{2}x^{-\frac{1}{3}}\)
\(18\ln 2 - \frac{27}{4}\) cao
A1
NB A0 for \(6\ln 8 - \frac{27}{4}\)
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Ax^{\frac{2}{3}}\ln x - \int Bx^{\frac{2}{3}} \times \frac{1}{x}\,dx$ | M1* | $A$ and $B$ are non-zero constants |
| $\frac{3}{2}x^{\frac{2}{3}}\ln x - \int\frac{3}{2}x^{\frac{2}{3}} \times \frac{1}{x}\,dx$ | A1 | ignore $+c$; NB $\frac{3}{2}x^{\frac{2}{3}}\ln x - \int\frac{3}{2}x^{-\frac{1}{3}}dx$; allow both marks if $dx$ omitted |
| $F[x] = \frac{3}{2}x^{\frac{2}{3}}\ln x - \frac{\frac{3}{2}}{\frac{2}{3}}x^{\frac{2}{3}}$ | A1 | ignore limits for first three marks |
| $F[8] - F[1]$ | M1*dep | dependent on integration of their $\frac{3}{2}x^{-\frac{1}{3}}$ |
| $18\ln 2 - \frac{27}{4}$ cao | A1 | NB A0 for $6\ln 8 - \frac{27}{4}$ |
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4 Find the exact value of $\int _ { 1 } ^ { 8 } \frac { 1 } { \sqrt [ 3 ] { x } } \ln x \mathrm {~d} x$, giving your answer in the form $A \ln 2 + B$, where $A$ and $B$ are constants to be found.
\hfill \mbox{\textit{OCR C4 2016 Q4 [5]}}