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

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

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

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

- 1 + x

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= x \cdot \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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

\frac{1}{x} - 1

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

Simplify

= - 1 + x

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