What is the integral from 1 to 2 of 4^x ?

The integral from 1 to 2 of 4^x is 6/(ln(2))

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integral from 1 to 2 of 4^x

\int_{1}^{2} 4^{x} d x

\frac{6}{ln ( 2 )}

Decimal Notation

8.65617 \dots

Solution steps

\int_{1}^{2} 4^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= [\frac{4 ^{x}}{ln ( 4 )}]_{1}^{2}

\frac{4 ^{x}}{ln ( 4 )} = \frac{2 ^{2 x - 1}}{ln ( 2 )}

= [\frac{2 ^{2 x - 1}}{ln ( 2 )}]_{1}^{2}

Compute the boundaries:

\frac{6}{ln ( 2 )}

= \frac{6}{ln ( 2 )}

4 x^{2}, \frac{x}{2}

x \to 0 - lim (- x + 3)

\int_{- 1}^{1} π (1 - x^{2}) d x

x \to \infty lim (1.003)

y^{^{'}} = x y^{3} (1 + x^{2})^{- \frac{1}{2}}

What is the integral from 1 to 2 of 4^x ?
The integral from 1 to 2 of 4^x is 6/(ln(2))