What is the integral from-3 to-2 of 1/x ?

The integral from-3 to-2 of 1/x is ln(2)-ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-3 to-2 of 1/x

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

ln (2) - ln (3)

Decimal Notation

- 0.40546 \dots

Solution steps

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{- 3}^{- 2}

Compute the boundaries:

ln (2) - ln (3)

= ln (2) - ln (3)

\int 0.5 e^{- 0.5 x} d x

\frac{d}{d x} (4 π)

5 s

\frac{c e ^{t}}{1 + c e ^{t}}

\frac{\partial}{\partial x} ((\frac{( x + 4 )}{y + 5})^{\frac{1}{2}})

What is the integral from-3 to-2 of 1/x ?
The integral from-3 to-2 of 1/x is ln(2)-ln(3)