What is the integral from 0 to 2 of 7^{-x} ?

The integral from 0 to 2 of 7^{-x} is (48)/(49ln(7))

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integral from 0 to 2 of 7^{-x}

\int_{0}^{2} 7^{- x} d x

\frac{48}{49 ln ( 7 )}

Decimal

0.50341 \dots

Solution steps

\int_{0}^{2} 7^{- x} d x

Apply u-substitution

: \int_{0}^{- 2} - 7^{u} d u

= \int_{0}^{- 2} - 7^{u} d u

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

= - \int_{- 2}^{0} - 7^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{0} 7^{u} d u)

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

= - (- [\frac{7 ^{u}}{ln ( 7 )}]_{- 2}^{0})

Simplify

= [\frac{7 ^{u}}{ln ( 7 )}]_{- 2}^{0}

Compute the boundaries:

\frac{48}{49 ln ( 7 )}

= \frac{48}{49 ln ( 7 )}

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What is the integral from 0 to 2 of 7^{-x} ?
The integral from 0 to 2 of 7^{-x} is (48)/(49ln(7))