What is the integral from 0 to 4 of 2^x ?

The integral from 0 to 4 of 2^x is (15)/(ln(2))

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integral from 0 to 4 of 2^x

\int_{0}^{4} 2^{x} d x

\frac{15}{ln ( 2 )}

Decimal Notation

21.64042 \dots

Solution steps

\int_{0}^{4} 2^{x} d x

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

= [\frac{2 ^{x}}{ln ( 2 )}]_{0}^{4}

Compute the boundaries:

\frac{15}{ln ( 2 )}

= \frac{15}{ln ( 2 )}

\frac{d}{d x} (2 x^{3} + 3)

\int e^{t} d t

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

x y^{^{'}} + y = x e^{x}

\frac{d}{d x} (c_{1} e^{- x} + c_{2} x e^{- x})

What is the integral from 0 to 4 of 2^x ?
The integral from 0 to 4 of 2^x is (15)/(ln(2))