To begin with, the best way I would choose to solve this is to find the present value in each of the three cases as:
$$PV = 190\times(\bar{a}_{12}+v^{12}\bar{a}_{8}) = 190\times(\dfrac{1-e^{-0.05\times12}}{0.05}+e^{-0.05\times12}\dfrac{1-e^{-\delta\times8}}{\delta})$$
Which gives, for $i=0.03$, $190\times12.9271$, for $i=0.05$, $190\times12.6424$, and for $i=0.07$, $190\times12.3856$. If you were to use the integration method as you did, then your formula should be:
$$PV=\int_0^{20}\rho(t)v(t)dt=190\times\int_0^{20}v(t)dt$$
Then $v(t)=e^{-0.05t}$ for $t<12$, and for $\delta=0.03$, $0.05$, or $0.07$, it's $e^{-0.05\times12}\times e^{-\delta (t-12)}$. Therefore, our formula becomes
$$PV=190\times(\int_0^{12}v(t)dt + e^{-0.6}\int_{12}^{20}v(t)dt)$$
$$=190\times(\int_0^{20}e^{-0.05t}dt + e^{-0.6}\int_{0}^{8}e^{-\delta t}dt)$$
Which yields the same values as above.
I've put in full working just in case, but your mistake seems to be integrating from $12$ to $t$ instead of from $12$ to $20$, which would give you $\delta(t-12)$ as I've shown above (instead of $8\delta$). As a reference check, remember that $v(t)$ should (almost) always vary with $t$, since it's the present value of 1 at time $t$. The final answer should be £2403.38.
I believe an easy way to understand the Kelly's criterion is the following:
Think about the following growth process: you have a discrete-time process where in each step you invest and amount/proportion $f$ of your current capital, and you start with an amount of cash equal to $W_0$. Each time-step, if you win, your wealth will increase by the product $(1-f)+f(1+b) = (1+fb)$ with $b>0$ being equal to the return rate each time you win, and conversely, if you loose at the actual stage your wealth will decrease by the product $(1-f)+f(1-a)=(1-fa)$ with $a>0$ being equal to the return rate reduction each time you loose (so, if my return is $-6\%$ the reduction in my wealth is going to be given by the product $(1-f\cdot 6\%)$). All of these, assuming that your probability of winning or loosing at each stage are independent, lets say, you could be winning at each stage with probability $p$ or loosing with probability $q = 1-p$.
With these, note that the order of winning and looses don't matter, so my current wealth at any stage $t=T$ will be given by the number of winning $N$ and the number of looses $M$ (where $N+M = T$):
$$ \begin{array}{r c l}
W_t & = & W_0 \cdot (1+fb) \cdot (1+fb) \cdot (1-fa) \cdots (1-fa) \cdot (1+fb) \\
& = & W_0\cdot(1+fb)^N \cdot (1-fa)^M \end{array}$$
Now, let try to calculate the equivalent Compound Growth rate $\tau$ for the process at stage $T$
$$\begin{array}{rcl}
(1+\tau)^T & = & (1+fb)^N \cdot (1-fa)^M \\
& = & \left( (1+fb)^\frac{N}{T} \cdot (1-fa)^\frac{M}{T}\right)^T \\
& = & \left( (1+fb)^\frac{N}{N+M} \cdot (1-fa)^\frac{M}{N+M}\right)^T \\
& \overset{\text{behave as when}\,T\to\infty}{\approx} & \left( (1+fb)^p \cdot (1-fa)^q\right)^T \\
\end{array}$$
so if you think as the time-expected-average path of the process could be defined as $\left< W_t \right>_t = W_0 \cdot (1+\tau)^t$ you can now figure out the meaning of the proportion $1+\tau = (1+fb)^p \cdot (1-fa)^q$ which is maximized through the Kelly's proportion $f^* = \frac{pb-aq}{ab}$, and understanding also the relation of the exponents that will tend to be the winning/loosing probabilities at each stage.
I believe that John Larry Kelly Jr. made one of the most important discoveries in Information Theory through this, founding that exist an optimal investing proportion which make a random process to growth exponentially, and since it doesn´t arise directly from Markowitz Theory it could be saying that important improvements could be done to optimal portfolios theories (maybe, maximizing the mode instead of the mean, but it just speculation right now). An interesting review is given by Edward O. Thorp here.
Additional review:
Think about the required return at each stage (betting my whole wallet each time): if I am betting in the casino, at each bet I could loose the whole amount I am investing so $1-a=100\%$, so, if each bet gives a prize of $\nu = 1+b$, at each bet I will win or loose with an outcome given by the expected value $E_t = (\nu p -1\cdot(1-p)) > 1$ if I want to be making money in the long term, so it requires that $\nu > 2/p-1$ which is always $\nu >1$ and highly restrictive, as example, for $p=0.5$ you need a prize of at least a $\nu >3$ times your bet.
But now, thinking in investing, if I loose (go bankrupt), I still can sell my assets to pay the doubt and keep some of my money (if I was and responsible manager), so the loosing proportion $\eta = (1-a)$ will be less than one (different from making a bet), so now the outcome given by the expected value $E_t = \nu p -\eta(1-p) > 1$ if I want to be making money in the long term, so will require that $\nu > 1/p+\eta(1-p)/p$, which could be even $\nu < 1$ and having an $E_t > 1$ given some values from $\eta$, so is sensefull to be doing investing, and is also different from making bets.
Right now coming back to Kelly's, if I invest a proportion $f$ on each trial/stage, the expected outcome by stage will be $E_t = f(\nu p -\eta(1-p)) > 1$ to be winning in the long term, so it will require that $f<1/(p(b-a)+2p+a-1)$ so doing bets as martingales are actually bad strategies even if I have residual value in my investing.
Best Answer
If the increasing rate and decreasing rate are $2\%$ then your factors $1.02$ and $0.98$ are right. If you have $n$ trials, then you have $\binom{n}{x}$ ways of making $x$ times profit and $n-x$ times making loss. The probability of making profit is $p$ and therefore the probability of making loss is $1-p$
Then the probability of making profit $x$ times and making $n-x$ times loss (in exactly that order) is $p^x\cdot (1-p)^{n-x}$.
The outcome after making profit $x$ and making $n-x$ times loss (in exactly that order) is $1.02^x\cdot 0.98^{n-x}$
The number of ways making profit $x$ times and making $n-x$ times loss is $\binom{n}x$. Let $EAR$ be the expected account ratio. If $EAR=1$, the winning/loosing is $0$
Then $(EAR)^n$ after $n$ trials is
$$(EAR)^n=\sum\limits_{x=0}^n \binom{n}x 1.02^x\cdot 0.98^{n-x}\cdot p^x\cdot (1-p)^{n-x}$$
$$=\sum\limits_{x=0}^n \binom{n}x (1.02p)^x\cdot (0.98\cdot (1-p))^{n-x}$$
The binomial theorem is $ (a+b)^n=\sum\limits_{x=0}^{n}{{n \choose x}\cdot a^{x}\cdot b^{n-x}} $. If we apply that we get
$$(EAR)^n=(1.02p+0.98\cdot (1-p))^n$$
Now we calculate $EAR$ by using the geometric mean
$$EAR=1.02p+0.98\cdot (1-p)$$
If $p=0.5$, then $EAR=1$. This is not surprisingly, since the possible amounts of winning and loosing are equal at a given level.