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

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

= 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

\int_{- 2}^{1} \frac{1}{x ^{3}} d x

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

\int_{2}^{4} \frac{1}{8} (6 - x - y) d y

\int_{0}^{1} \frac{1}{e ^{2 x}} d x

\int_{- 3}^{9} 3 x - 9 d x

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