Area under decay curve exp(-0.6969t/h) where t is the time (with t=0 initially) and h is a constant "half life" is analytically integrable, but what if the half life is increasing with time? I. e. if h(t) = H + at. (Note exp(-0.6969) is not exactly 0.5 but close and easy to remember.) This problem arises when trying to evaluate the total blocking of a "puff" green house gas, like methane CH4 released at t = 0. CH4 in the air is mainly destroyed by the OH- radical. OH- production rate is limited by the flux of harsh solar UV. For at least the 800,000 years this production rate was greater than the release of CH4, but that is no longer true. http://www.epa.gov/climatechange/images/indicator_figures/ghg-concentrations-figure2-2014.png For the last few decades, the CH4 release rate has been faster than the OH- is formed, as seen above. I.e. now the CH4 is destroying the OH- so the half life of CH4 is increasing. The CH4 half life was 9.6 years in 2003 and 12.4 in 2013*. Lets assume half life is given by: h(t) = 12.6(1+ 0.3t) with t in years and t = 0 on 1/1/2015. Can you integrate how many "puff years" exist for some specific period of years? As CO2 has a half life of more than 1000 years, other green house gas puffs are usually compare to it. I.e. How much more global warming a puff of CH4 will make compared to an equal mass of CO2 released at the same time, during the next X years. Here are some results: t = 1/365: 120 times more than the CO2 did. t = 10: 104 times more t = 20: 83.8 times more t = 50: 48.4 times more t = 100: 28.5 times more t = 500: 8.1 times more So little of the puff's CH4 is left after 1000 years probably is less than one as about half of a fossil fuel released puff would still exist. Some of the comparison puff of CO2 molecules may have spent many decades absorbed in the ocean, before re-entering the air; but not all will be part of shells lying on the bottom, etc. If you can give results for some of the same year with above h(t) I thank you. Please also give integral for h = 12.4 as that was assumed for data above. With that and the above data I can convert your results into a new table for the years you give both. * Some report it as 12.6 in 2013.