You invest 100 dollars with continuously compounding interest. The interest rate is compounded annually with interest rate $R$ that has a pdf of $f(r)= 1.5r - 0.75r^2$ 0<=r<=2. Lets say the money is kept for $T$ years where $T$ is random variable, exponentially distributed with parameter $\lambda$. As a function of $\lambda$, what is the expected value of the account after $T$ years
I know that I need to find $E[100(1+R)^T]$ but not sure how to come up with the joint pdf of $R$ and $T$.
You should be computing $\mathbb{E}[e^{RT}]$ if you are interested in continuously compounded interest. If $R$ and $T$ are independent and $\lambda>2$ then this quantity is \begin{multline*} \int_{0}^{2}\int_{0}^{\infty}f_{R}(r)f_{T}(t)e^{rt}drdt=\int_{0}^{2}f_{R}(r)\int_{0}^{\infty}\lambda e^{(r-\lambda)t}dtdr\\ \int_{0}^{2}\left(1.5r-0.75r^{2}\right)\frac{\lambda}{\lambda-r}dr=\frac{3\lambda}{4}\left(2\left(\lambda-1\right)+\left(\lambda-2\right)\lambda\log\left(\frac{\lambda-2}{\lambda}\right)\right) \end{multline*}
The condition $\lambda > 2$ is needed to ensure integrability. You can check that as $\lambda \downarrow 2$, you become an "infillionaire".