75 Integration

$$|\lambda f(x)+\mu g(x)|\le M{e}^{\beta x},$$

so we can integrate

$${\int }_0^{\mathrm{\infty }}{e}^{-sx}(\lambda f(x)+\mu g(x))𝑑x=\lambda {\int }_0^{\mathrm{\infty }}{e}^{-sx}f(x)𝑑x+\mu {\int }_0^{\mathrm{\infty }}{e}^{-sx}g(x)𝑑x.$$

(iv) Suppose that $|f^{\prime }(x)|\le M{e}^{\beta x}$ for all $x>0$ and some $\beta >0$. Then

$$f(x)=f(0)+{\int }_0^x{f}^{\prime }(t)𝑑t,$$

so

$$|f(x)|\le |f(0)|+{\int }_0^{x}M{e}^{\beta t}𝑑t$$

$$=|f(0)|+{\left[\frac{M{e}^{\beta t}}{\beta }\right]}_0^x$$