What is the integral from 0 to 2 of 3^{-x} ?

The integral from 0 to 2 of 3^{-x} is 8/(9ln(3))

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integral from 0 to 2 of 3^{-x}

\int_{0}^{2} 3^{- x} d x

\frac{8}{9 ln ( 3 )}

Decimal

0.80910 \dots

Solution steps

\int_{0}^{2} 3^{- x} d x

Apply u-substitution

: \int_{0}^{- 2} - 3^{u} d u

= \int_{0}^{- 2} - 3^{u} d u

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

= - \int_{- 2}^{0} - 3^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{0} 3^{u} d u)

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

= - (- [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{0})

Simplify

= [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{0}

Compute the boundaries:

\frac{8}{9 ln ( 3 )}

= \frac{8}{9 ln ( 3 )}

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{x} cos (3 x)

\int sin (u^{2}) d u

\int \frac{37}{x ^{3} - 27} d x

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

\frac{x ^{2}}{8 + x}

What is the integral from 0 to 2 of 3^{-x} ?
The integral from 0 to 2 of 3^{-x} is 8/(9ln(3))