What is the integral from-1 to 1 of (e^{-3x}) ?

The integral from-1 to 1 of (e^{-3x}) is (e^6-1)/(3e^3)

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integral from-1 to 1 of (e^{-3x})

\int_{- 1}^{1} (e^{- 3 x}) d x

\frac{e ^{6} - 1}{3 e ^{3}}

Decimal

6.67858 \dots

Solution steps

\int_{- 1}^{1} e^{- 3 x} d x

Apply u-substitution

: \int_{3}^{- 3} - \frac{1}{3} e^{u} d u

= \int_{3}^{- 3} - \frac{1}{3} e^{u} d u

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

= - \int_{- 3}^{3} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 3}^{3} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 3}^{3})

Simplify

= \frac{1}{3} [e^{u}]_{- 3}^{3}

Compute the boundaries:

e^{3} - \frac{1}{e ^{3}}

= \frac{1}{3} (e^{3} - \frac{1}{e ^{3}})

Simplify

= \frac{e ^{6} - 1}{3 e ^{3}}

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What is the integral from-1 to 1 of (e^{-3x}) ?
The integral from-1 to 1 of (e^{-3x}) is (e^6-1)/(3e^3)