What is the integral from-1 to 2 of-2e^{1-3x} ?

The integral from-1 to 2 of-2e^{1-3x} is -(2(e^9-1))/(3e^5)

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integral from-1 to 2 of-2e^{1-3x}

\int_{- 1}^{2} - 2 e^{1 - 3 x} d x

- \frac{2 ( e ^{9} - 1 )}{3 e ^{5}}

Decimal Notation

- 36.39427 \dots

Solution steps

\int_{- 1}^{2} - 2 e^{1 - 3 x} d x

Take the constant out:

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

= - 2 \cdot \int_{- 1}^{2} e^{1 - 3 x} d x

Apply u-substitution

: \int_{4}^{- 5} - \frac{1}{3} e^{u} d u

= - 2 \cdot \int_{4}^{- 5} - \frac{1}{3} e^{u} d u

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

= - 2 (- \int_{- 5}^{4} - \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

= - 2 (- (- \frac{1}{3} \cdot \int_{- 5}^{4} e^{u} d u))

Use the common integral:

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

= - 2 (- (- \frac{1}{3} [e^{u}]_{- 5}^{4}))

Simplify

- 2 (- (- \frac{1}{3} [e^{u}]_{- 5}^{4})) : - \frac{2}{3} [e^{u}]_{- 5}^{4}

= - \frac{2}{3} [e^{u}]_{- 5}^{4}

Compute the boundaries:

e^{4} - \frac{1}{e ^{5}}

= - \frac{2}{3} (e^{4} - \frac{1}{e ^{5}})

Simplify

= - \frac{2 ( e ^{9} - 1 )}{3 e ^{5}}

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What is the integral from-1 to 2 of-2e^{1-3x} ?
The integral from-1 to 2 of-2e^{1-3x} is -(2(e^9-1))/(3e^5)