Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{{16}^x}}}{{{4^{2x}}}}dx} \cr
& {\text{write 16 as }}{{\text{4}}^2} \cr
& = \int_0^1 {\frac{{{{\left( {{4^2}} \right)}^x}}}{{{4^{2x}}}}} dx \cr
& {\text{using the property of the exponents }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& = \int_0^1 {\frac{{{4^{2x}}}}{{{4^{2x}}}}} dx \cr
& {\text{Simplify}} \cr
& = \int_0^1 {dx} \cr
& {\text{integrate and evaluate}} \cr
& = \left( x \right)_0^1 \cr
& = 1 - 0 \cr
& = 1 \cr} $$
