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

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

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integral from-infinity to 1 of 2^x

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

\frac{2}{ln ( 2 )}

Decimal

2.88539 \dots

Solution steps

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

Compute the indefinite integral:

\int 2^{x} d x = \frac{2 ^{x}}{ln ( 2 )} + C

Compute the boundaries:

\int_{- \infty}^{1} 2^{x} d x = \frac{2}{ln ( 2 )} - 0

= \frac{2}{ln ( 2 )} - 0

Simplify

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

\int x (x + 1)^{9} d x

\int 2 2 x - 1 d x

\int \frac{5}{( 3 x - 1 ) ^{5}} d x

\frac{\partial}{\partial x} (4 x^{2} y^{3} - 4)

y^{^{'}} = 0.2 ln (\frac{3000}{y}) y, y (0) = 1000

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