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

The integral from 0 to 1 of (e^{-2x})/2 is (e^2-1)/(4e^2)

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

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

\frac{e ^{2} - 1}{4 e ^{2}}

Decimal

0.21616 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{2} \cdot \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

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

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

Take the constant out:

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

= \frac{1}{2} (- (- \frac{1}{2} \cdot \int_{- 2}^{0} e^{u} d u))

Use the common integral:

\int e^{u} d u = e^{u}

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

Simplify

\frac{1}{2} (- (- \frac{1}{2} [e^{u}]_{- 2}^{0})) : \frac{1}{4} [e^{u}]_{- 2}^{0}

= \frac{1}{4} [e^{u}]_{- 2}^{0}

Compute the boundaries:

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

= \frac{1}{4} (1 - \frac{1}{e ^{2}})

Simplify

= \frac{e ^{2} - 1}{4 e ^{2}}

What is the integral from 0 to 1 of (e^{-2x})/2 ?
The integral from 0 to 1 of (e^{-2x})/2 is (e^2-1)/(4e^2)