# Evaluate the following integrals. int_0^1frac{16^x}{4^{2x}}dx

Evaluate the following integrals.
${\int }_{0}^{1}\frac{{16}^{x}}{{4}^{2x}}dx$
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$\text{Solution:}$
${\int }_{0}^{1}\frac{{16}^{x}}{{4}^{2x}}dx$
$\text{Om simplifying further we get:}$
${\int }_{0}^{1}\frac{{16}^{x}}{{4}^{2x}}dx={\int }_{0}^{1}\frac{\left({4}^{2}{\right)}^{x}}{{4}^{2x}}dx$
$={\int }_{0}^{1}\frac{{4}^{2x}}{{4}^{2x}}dx$
$={\int }_{0}^{1}1dx$
$=\left[x{\right]}_{0}^{1}$
$=\left[1-0\right]$
$=1$