Answer
$$f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 0.7{e^{ - 0.7x}};\,\,\,\,\,\,\,\left[ {0,\infty } \right) \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right]. \cr
& 0.7{e^{ - 0.7x}}{\text{ is positive for all real numbers }}x \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_0^\infty {0.7{e^{ - 0.7x}}} dx \cr
& {\text{solve the improper integral using the definition }}\int_a^\infty {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_a^b {f\left( x \right)} dx{\text{ }} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_0^b {0.7{e^{ - 0.7x}}} dx \cr
& {\text{integrating}} \cr
& - \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 0.7x}}} \right]_0^b \cr
& - \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 0.7b}} - {e^0}} \right] \cr
& - \mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - 0.7b}} - 1} \right] \cr
& {\text{evaluate the limit when b}} \to \infty \cr
& - \left( {{e^{ - \infty }} - 1} \right) \cr
& {\text{simplifying}} \cr
& - \left( {0 - 1} \right) \cr
& = 1 \cr
& \cr
& {\text{The condition 1 and 2 are verified}}{\text{, so }} \cr
& f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
