What is the integral from-1 to 1 of 12e^{-3x} ?

The integral from-1 to 1 of 12e^{-3x} is 4(e^3-1/(e^3))

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integral from-1 to 1 of 12e^{-3x}

\int_{- 1}^{1} 12 e^{- 3 x} d x

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

Decimal

80.14299 \dots

Solution steps

\int_{- 1}^{1} 12 e^{- 3 x} d x

Take the constant out:

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

= 12 \cdot \int_{- 1}^{1} e^{- 3 x} d x

Apply u-substitution

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

= 12 \cdot \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

= 12 (- \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

= 12 (- (- \frac{1}{3} \cdot \int_{- 3}^{3} e^{u} d u))

Use the common integral:

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

= 12 (- (- \frac{1}{3} [e^{u}]_{- 3}^{3}))

Simplify

12 (- (- \frac{1}{3} [e^{u}]_{- 3}^{3})) : 4 [e^{u}]_{- 3}^{3}

= 4 [e^{u}]_{- 3}^{3}

Compute the boundaries:

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

= 4 (e^{3} - \frac{1}{e ^{3}})

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What is the integral from-1 to 1 of 12e^{-3x} ?
The integral from-1 to 1 of 12e^{-3x} is 4(e^3-1/(e^3))