Standard +0.3 This is a straightforward integration problem requiring standard exponential integration rules and algebraic manipulation to solve for k. While it involves multiple exponential terms and requires careful arithmetic with the limits, it's a routine C3 exercise with no conceptual difficulty—slightly easier than average due to its mechanical nature.
Integrate $e^{3x}$ to obtain $\frac{1}{3}e^{3x}$ or $e^{-\frac{1}{k}x}$ to obtain $-2e^{-\frac{1}{k}x}$ | B1 | or both
Obtain indefinite integral of form $m_1e^{3x}+m_2e^{-\frac{1}{k}x}$ | M1 | any constants $m_1$ and $m_2$
Obtain correct $\frac{1}{3}e^{3x}-2(k-2)e^{-\frac{1}{k}x}$ | A1 | or equiv
Obtain $e^{\ln 4}=64$ or $e^{-\frac{1}{k}\ln 4}=\frac{1}{4}$ | B1 | or both
Apply limits and equate to 185 | M1 | including substitution of lower limit
Obtain $\frac{64}{3}-k-(k-2)-\frac{1}{4}k+2(k-2)=185$ | A1 | or equiv
Obtain $\frac{63}{2}$ | A1 | 7 or equiv
**Total: 7**
---
Given that
$$\int_0^{\ln 4} (ke^{3x} + (k - 2)e^{-\frac{x}{3}}) \, dx = 185,$$
find the value of the constant $k$. [7]
\hfill \mbox{\textit{OCR C3 2010 Q6 [7]}}