What is the integral from 0 to ln(x) of e^{-y} ?

The integral from 0 to ln(x) of e^{-y} is -1/x+1

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integral from 0 to ln(x) of e^{-y}

\int_{0}^{l n (x)} e^{- y} d y

- \frac{1}{x} + 1

Solution steps

\int_{0}^{l n (x)} e^{- y} d y

Apply u-substitution

: \int_{0}^{- l n (x)} - e^{u} d u

= \int_{0}^{- l n (x)} - e^{u} d u

Take the constant out:

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

= - \int_{0}^{- l n (x)} e^{u} d u

Use the common integral:

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

= - [e^{u}]_{0}^{- l n (x)}

Compute the boundaries:

\frac{1}{x} - 1

= - (\frac{1}{x} - 1)

Simplify

= - \frac{1}{x} + 1

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What is the integral from 0 to ln(x) of e^{-y} ?
The integral from 0 to ln(x) of e^{-y} is -1/x+1