Since your temperature readings aren't equispaced, you can't directly apply Simpson's rule; the approach equivalent to this is to construct the parabola that interpolates three consecutive points (i.e., across two panels), and integrate that. The problem with this approach, of course, is that you need to have an odd number of data points (even number of panels) to do this.
You can use the trapezoidal rule, of course, across each panel, but a probably better idea might be to construct a cubic Hermite interpolant for each panel, and then integrate that. An obvious problem is that four conditions are needed to uniquely determine a cubic for each panel (two points and two derivative values), but estimates of derivative values can be constructed from the data such that the piecewise interpolant is locally monotonic; briefly, an piecewise interpolant is locally monotonic if there are no spurious inflection points or extrema within a panel. One approach to estimating derivative values for monotonic interpolation, due to Fritsch and Carlson, is implemented in FORTRAN as the pchip package, and in MATLAB as the function pchip. A more modern approach, and one which may be better is some situations, is due to Steffen. You may have to experiment which of the trapezoidal rule, Fritsch-Carlson, or Steffen would be best for integrating your data.
Let me detail the way one uses cubic Hermite interpolation for integrating data:
Each panel is bounded by two points, $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$. The trapezoidal rule consists of constructing the line joining these two points (linear interpolation):
$$f_i(x)=\frac{x_{i+1}-x}{x_{i+1}-x_i}y_i+\frac{x-x_i}{x_{i+1}-x_i}y_{i+1}$$
and integrating that:
$$\int_{x_i}^{x_{i+1}}f_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{2}(y_i+y_{i+1})$$
Integrating with cubic Hermite interpolation can be considered as a further "improvement" of the trapezoidal rule; briefly, in addition to points, one has derivative values $y_i^{\prime}$ and $y_{i+1}^{\prime}$, from which one constructs the cubic
$$g_i(x)=y_i+y_i^{\prime}(x-x_i)+c_i(x-x_i)^2+d_i(x-x_i)^3$$
where
$\displaystyle c_i=\frac{3\frac{y_{i+1}-y_i}{x_{i+1}-x_i}-2y_i^{\prime}-y_{i+1}^{\prime}}{x_{i+1}-x_i}$ and $\displaystyle d_i=\frac{y_i^{\prime}+y_{i+1}^{\prime}-2\frac{y_{i+1}-y_i}{x_{i+1}-x_i}}{(x_{i+1}-x_i)^2}$.
Integrating $g_i(x)$ might look slightly complicated, however, there is a nice expression for the integral:
$$\int_{x_i}^{x_{i+1}}g_i(x)\mathrm dx=\frac{x_{i+1}-x_i}{6}\left(y_i+4g_i\left(\frac{x_{i+1}+x_i}{2}\right)+y_{i+1}\right)$$
whose verification I'll leave up to you.
I'll just add the note that in the case of equispaced data, integration with a piecewise cubic Hermite interpolant is equivalent to integration with the trapezoidal rule plus corrections at the beginning and end.
Best Answer
Notice that
\begin{align*} \int_{12}^{\infty} \frac{1}{te^t} \, dt &\le \sum_{n = 12}^{\infty} \frac{1}{ne^n} \\ &\le \frac 1 {12e^{12}} \sum_{n = 0}^{\infty} \frac 1 {e^n} \\ &< \frac{1}{e^{12}} \\ &< 0.000006 \end{align*}
is already small enough that it won't affect your computation's first four decimal places. Now just compute $\int_{1/2}^{12}$ using your favorite numerical method. (And by the way, this was a very crude method of estimate, so $12$ is a very loose upper bound.)