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

The integral from-1 to 1 of 2^x is 3/(2ln(2))

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integral from-1 to 1 of 2^x

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

\frac{3}{2 ln ( 2 )}

Decimal Notation

2.16404 \dots

Solution steps

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

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

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

Compute the boundaries:

\frac{3}{2 ln ( 2 )}

= \frac{3}{2 ln ( 2 )}

\frac{d}{d x} (2 + \frac{3}{1 + ( x + 1 ) ^{2}})

\frac{d}{d x} (8 x sin (x))

x \to - 3 + lim (\frac{x - 3}{x - 4})

\frac{d}{d x} (\frac{e ^{x^{2}}}{( 2 x + 1 ) ^{5}})

\frac{\partial}{\partial y} (4 x y^{3})

What is the integral from-1 to 1 of 2^x ?
The integral from-1 to 1 of 2^x is 3/(2ln(2))