What is the integral from 0 to 1 of (2^x-2) ?

The integral from 0 to 1 of (2^x-2) is 1/(ln(2))-2

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integral from 0 to 1 of (2^x-2)

\int_{0}^{1} (2^{x} - 2) d x

\frac{1}{ln ( 2 )} - 2

Solution steps

\int_{0}^{1} 2^{x} - 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 2^{x} d x - \int_{0}^{1} 2 d x

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

\int_{0}^{1} 2 d x = 2

= \frac{1}{ln ( 2 )} - 2

\int_{- 1}^{0} 384 x^{5} (8 x^{6} - 3)^{3} d x

\int_{0}^{1} (2 x + 1)^{5} d x

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\int_{- 1}^{1} k (1 - x) d x

\int_{0}^{1} (x^{3} + 4 x) d x

What is the integral from 0 to 1 of (2^x-2) ?
The integral from 0 to 1 of (2^x-2) is 1/(ln(2))-2