Integral calculus exp

Question

I have the following integral $\int_0^\infty e^{-6t}~dt=\dfrac{1}{6}$ and I can't remember the properties of integrals or "$e$" to get the result.

Can you, please, help me, with the explanation?

Thank you!

Answer 1

Given that $e^y=\frac{de^y}{dy}$ for all $y \in \mathbb{R}$, we have: $$\int_{-\infty}^xe^{y}dy=\int_{-\infty}^xde^{y}=e^x- \lim_{y \to -\infty}e^y=e^x-0=e^x$$ So, by substituting $y=-6t$, we can use this propety. We calculate: $$dt=-\frac{1}{6}dy, 0 \to0,\infty \to - \infty$$ The integral then becomes: $$\int_0^\infty e^{-6t}dt=\int_0^{-\infty}-\frac{1}{6}e^{y}dy=\frac{1}{6}\int_{-\infty}^0e^{y}dy=\frac{1}{6}*e^0=\frac{1}{6}*1=\frac{1}{6}$$

Answer 2

Use the change of variable $z=-6t$.