Moderate -0.3 This is a straightforward reverse chain rule integration followed by solving an exponential equation. The integration is standard (e^(2x+1) with coefficient adjustment), and solving for 'a' requires basic logarithm manipulation. Slightly easier than average due to being a direct application of a single technique with no conceptual challenges.
Apply both limits correctly and rearrange at least to \(e^{2a+1} = \ldots\)
M1
Use logarithms correctly to find \(a\)
M1
Obtain 1.097
A1
[5]
Obtain integral of form $ke^{2x+1}$ | M1 |
Obtain correct $3e^{2x+1}$ | A1 |
Apply both limits correctly and rearrange at least to $e^{2a+1} = \ldots$ | M1 |
Use logarithms correctly to find $a$ | M1 |
Obtain 1.097 | A1 | [5]
5 Given that $\int _ { 0 } ^ { a } 6 \mathrm { e } ^ { 2 x + 1 } \mathrm {~d} x = 65$, find the value of $a$ correct to 3 decimal places.
\hfill \mbox{\textit{CAIE P2 2016 Q5 [5]}}