What is the integral from 0 to 1 of (1-e^2)^{1/2} ?

The integral from 0 to 1 of (1-e^2)^{1/2} is i(e^2-1)^{1/2}

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 (1-e^2)^{1/2}

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

i (e^{2} - 1)^{\frac{1}{2}}

Solution steps

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

Integral of a constant:

\int a d x = a x

= [(1 - e^{2})^{\frac{1}{2}} x]_{0}^{1}

Apply imaginary number rule:

- a = i a

= [i (- 1 + e^{2})^{\frac{1}{2}} x]_{0}^{1}

Compute the boundaries:

i (e^{2} - 1)^{\frac{1}{2}}

= i (e^{2} - 1)^{\frac{1}{2}}

\int_{0}^{2} (\frac{1}{1 + x ^{2}})^{2} d x

\int_{- π}^{π} (sin (x))^{2} d x

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

\int_{2}^{\infty} x^{- 4} d x

\int_{0}^{2} \frac{3}{8} (4 x - 2 x^{2}) d x

What is the integral from 0 to 1 of (1-e^2)^{1/2} ?
The integral from 0 to 1 of (1-e^2)^{1/2} is i(e^2-1)^{1/2}