Easy integral,density,definite integral

Question

I cannot compute the following integral. I'm confused with upper and lower bounds which are somehow switched from my point of view: $$\int_0^\infty\lambda e^{-\lambda x}dx=1.$$ This is supposed to be a normalized density of an exponential distribution. Can someone please guide me through this computation step by step so that I can follow it?

Answer 1

Note that :

$$(e^{-\lambda x})' = -\lambda e^{-\lambda x}$$

Thus, it is :

$$\lambda e^{-\lambda x} =-(e^{-\lambda x})'$$

The integral then becomes :

$$\int_0^\infty \lambda e^{-\lambda x} \mathrm{d}x = -\int_0^\infty \big(e^{-\lambda x}\big)'\mathrm{d}x = -\big[e^{-\lambda x}\big]_0^\infty$$

$$\implies$$

$$\int_0^\infty \lambda e^{-\lambda x} \mathrm{d}x = -\lim_{x \to \infty}e^{-\lambda x} + 1=1$$