What is the integral from 1 to 2 of (1/(x^2)) ?

The integral from 1 to 2 of (1/(x^2)) is 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of (1/(x^2))

\int_{1}^{2} (\frac{1}{x ^{2}}) d x

\frac{1}{2}

Solution steps

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

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{2}

= [- \frac{1}{x}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int - \frac{1}{1 + 2 x} d x

\frac{\partial}{\partial x} (e^{- y} sin (x))

\frac{d y}{d x} = \frac{2 x ^{3}}{y ^{2}}, y (2) = 3

\frac{d}{d x} (17 - x^{2} - 4 y^{2})

\int (4 - x^{2})^{3} d x

What is the integral from 1 to 2 of (1/(x^2)) ?
The integral from 1 to 2 of (1/(x^2)) is 1/2