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

Popular >

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

\int_{0}^{\frac{x}{4}} tan^{3} (x) d x

\int_{0}^{9} x e^{- \frac{x}{3}} d x

\int_{- 1}^{0} x^{3} - 6 x d x

\int_{1}^{4} \frac{7}{x ln ( 7 x )} d x

\int_{- 2}^{0} - x^{3} + 4 x d x

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