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

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

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

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

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

Decimal Notation

3.19452 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

e^{2} - 1

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

Simplify

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

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