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}

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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}}

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}