What is the integral from 0 to 1 of sqrt(x^2-1) ?

The integral from 0 to 1 of sqrt(x^2-1) is -Does not exist

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of sqrt(x^2-1)

\int_{0}^{1} x^{2} - 1 d x

- Does not exist

Solution steps

\int_{0}^{1} x^{2} - 1 d x

Compute the indefinite integral:

\int x^{2} - 1 d x = \frac{1}{2} x x^{2} - 1 - \frac{1}{2} ln x + x^{2} - 1 + C

Compute the boundaries:

\int_{0}^{1} x^{2} - 1 d x = 0 -

Does not exist

= 0 - Does not exist

Simplify

= - Does not exist

\int_{0}^{4} ln (3 t + 1) d t

\int_{0}^{2} (x ln (x)) d x

\int_{1}^{\infty} \frac{8}{7 x - 1} d x

\int_{- \infty}^{0} x e^{- 5 x} d x

\int_{0}^{t} e^{- 2 x} sin (x) d x

What is the integral from 0 to 1 of sqrt(x^2-1) ?
The integral from 0 to 1 of sqrt(x^2-1) is -Does not exist