What is the integral from 0 to 2 of 2^{-x} ?

The integral from 0 to 2 of 2^{-x} is 3/(4ln(2))

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integral from 0 to 2 of 2^{-x}

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

\frac{3}{4 ln ( 2 )}

Decimal

1.08202 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 2} - 2^{u} d u

= \int_{0}^{- 2} - 2^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{0} - 2^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 2}^{0} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0}

Compute the boundaries:

\frac{3}{4 ln ( 2 )}

= \frac{3}{4 ln ( 2 )}

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y})

y^{2} y^{^{'}} = y^{3} - 3 x, y (0) = 2

x \to 0 lim (2 (1 + 3 x)^{\frac{2}{x}})

x \to 8 lim (x^{\frac{4}{3}})

\frac{d}{d x} (\frac{3 x ^{2} + 2 x}{4 x - 2})

What is the integral from 0 to 2 of 2^{-x} ?
The integral from 0 to 2 of 2^{-x} is 3/(4ln(2))