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

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

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integral from 0 to 1 of (e^{-x})^2

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

\frac{1}{2} - \frac{1}{2 e ^{2}}

Decimal Notation

0.43233 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{\frac{1}{e}} - u d u

= \int_{1}^{\frac{1}{e}} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{e}}^{1} - u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{1}{e}}^{1} u d u)

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{\frac{1}{e}}^{1}

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

Simplify

= [\frac{u ^{2}}{2}]_{\frac{1}{e}}^{1}

Compute the boundaries:

\frac{1}{2} - \frac{1}{2 e ^{2}}

= \frac{1}{2} - \frac{1}{2 e ^{2}}

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