What is the integral from ln(2) to ln(3) of e^{-x} ?

The integral from ln(2) to ln(3) of e^{-x} is 1/6

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integral from ln(2) to ln(3) of e^{-x}

\int_{l n (2)}^{l n (3)} e^{- x} d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{l n (2)}^{l n (3)} e^{- x} d x

Apply u-substitution

= \int_{- l n (2)}^{- l n (3)} - e^{u} d u

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

= - \int_{- l n (3)}^{- l n (2)} - e^{u} d u

Take the constant out:

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

= - (- \int_{- l n (3)}^{- l n (2)} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- [e^{u}]_{- l n (3)}^{- l n (2)})

Simplify

= [e^{u}]_{- l n (3)}^{- l n (2)}

Compute the boundaries:

\frac{1}{6}

= \frac{1}{6}

What is the integral from ln(2) to ln(3) of e^{-x} ?
The integral from ln(2) to ln(3) of e^{-x} is 1/6