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I need help with the last step in the solution of this ODE \begin{equation} T'(t)-\lambda kT(t)=0. \end{equation} The integrating factor is $e^{-\int_{0}^{t} \lambda k \mathrm{d}s}=e^{-\lambda k t} $ so \begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\Big (T(t)e^{-\lambda k t}\Big)=0. \end{equation} Integration over $(0,t)$ so \begin{align} \int_{0}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Big(T(s)e^{-\lambda k s}\Big)\mathrm{d}t&=T(t)e^{-\lambda k t}-T(0)e^{-\lambda k 0}=0 \\ T(t)&=T(0)e^{\lambda k t} \end{align} How is the integral evaluated?

In the integral $T$ is now a function of $s$ but the integration variable is $t$, shouldn't it be \begin{align} \int_{0}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Big(T(s)e^{-\lambda k s}\Big)\mathrm{d}t&=\frac{\mathrm{d}}{\mathrm{d}s}\Big(T(s)e^{-\lambda k s}\Big)\int_{0}^{t}\mathrm{d}t\\ &=\frac{\mathrm{d}}{\mathrm{d}s} T(s)e^{-\lambda k s}\Big [ t \Big ]^t_{0}\\ &=\frac{\mathrm{d}}{\mathrm{d}s} T(s)e^{-\lambda k s}t \quad \text{?} \end{align}

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    This is a typo, it should be $ds$ in the integral, not $dt$. You use the fundamental theorem of calculus which states that $\int_0^t F'(s) \, ds = F(t) - F(0)$. – 2017-01-17

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