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

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

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

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

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

Solution steps

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

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

= \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}}

\int_{- 2}^{2} \frac{1}{2} (16 - x^{4}) d x

\int cos (3 x + 4) d x

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\frac{d}{d x} (\frac{a}{x ^{b}})

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