Moderate -0.3 This is a straightforward application of logarithmic integration using reverse chain rule, followed by solving a simple equation involving natural logarithms. The integral is standard (ln|2x-1|/2), and solving ln(2k-1)/ln(1) = 1 for k requires only basic log manipulation. Slightly easier than average due to being a single-technique problem with no conceptual challenges.
Obtain indefinite integral of the form \(a\ln(2x-1)\), where \(a = -\frac{1}{2}, 1,\) or \(2\)
M1
Use limits and obtain equation \(\frac{1}{2}\ln(2k-1) = 1\)
A1
Use correct method for solving an equation of the form \(a\ln(2k-1) = 1\), where \(a = -\frac{1}{2}, 1,\) or \(2\), for \(k\)
M1
Obtain answer \(k = \frac{1}{2}(e^2+1)\), or exact equivalent
A1
[4]
Obtain indefinite integral of the form $a\ln(2x-1)$, where $a = -\frac{1}{2}, 1,$ or $2$ | M1 |
Use limits and obtain equation $\frac{1}{2}\ln(2k-1) = 1$ | A1 |
Use correct method for solving an equation of the form $a\ln(2k-1) = 1$, where $a = -\frac{1}{2}, 1,$ or $2$, for $k$ | M1 |
Obtain answer $k = \frac{1}{2}(e^2+1)$, or exact equivalent | A1 | [4]
1 Find the exact value of the constant $k$ for which $\int _ { 1 } ^ { k } \frac { 1 } { 2 x - 1 } \mathrm {~d} x = 1$.
\hfill \mbox{\textit{CAIE P3 2007 Q1 [4]}}