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

Question
Integrals
Evaluate the following integrals.
$$\int_0^1\frac{16^x}{4^{2x}}dx$$

2021-03-10
$$\text{To integrate: }\int_0^1\frac{16^x}{4^{2x}}dx$$
$$\text{Solution:}$$
$$\int_0^1\frac{16^x}{4^{2x}}dx$$
$$\text{Om simplifying further we get:}$$
$$\int_0^1\frac{16^x}{4^{2x}}dx=\int_0^1\frac{(4^2)^x}{4^{2x}}dx$$
$$=\int_0^1\frac{4^{2x}}{4^{2x}}dx$$
$$=\int_0^1 1dx$$
$$=[x]_0^1$$
$$=[1-0]$$
$$=1$$
$$\textbf{Result: }\int_0^1\frac{16^x}{4^{2x}}dx=1$$

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