What is the integral from-5 to 0 of 2^{1-x} ?

The integral from-5 to 0 of 2^{1-x} is (62)/(ln(2))

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integral from-5 to 0 of 2^{1-x}

\int_{- 5}^{0} 2^{1 - x} d x

\frac{62}{ln ( 2 )}

Decimal

89.44709 \dots

Solution steps

\int_{- 5}^{0} 2^{1 - x} d x

Apply u-substitution

= \int_{6}^{1} - 2^{u} d u

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

= - \int_{1}^{6} - 2^{u} d u

Take the constant out:

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

= - (- \int_{1}^{6} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{6})

Simplify

= [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{6}

Compute the boundaries:

\frac{62}{ln ( 2 )}

= \frac{62}{ln ( 2 )}

What is the integral from-5 to 0 of 2^{1-x} ?
The integral from-5 to 0 of 2^{1-x} is (62)/(ln(2))