differential and integeral calculus

[MATH]T - T_{o} = 5\int_{0}^{30} te^{-0.1t} \ dt[/MATH]
where [MATH]T_{o}[/MATH] is the initial temperature and [MATH]T[/MATH] is the temperature after [MATH]30 \ s[/MATH]
Can you continue?
 
Thank you yes I can solve it now I just couldn't see how to get started thanks again
 
[MATH]400.42586326 \ K+ 288 \ K[/MATH]
And if you want to convert to Celsius, do this

If [MATH]288 \ K = 15^{o}C[/math]
This means [MATH]288 \ K - 273.15 = 14.85^{o}C \approx 15^{o}C[/MATH]
 
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