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

The integral from-2 to 3 of (3e^{-2x-1}) is (3(e^{10}-1))/(2e^7)

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

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

\frac{3 ( e ^{10} - 1 )}{2 e ^{7}}

Decimal Notation

30.12693 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{3}^{- 7} - \frac{1}{2} e^{u} d u

= 3 \cdot \int_{3}^{- 7} - \frac{1}{2} e^{u} d u

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

= 3 (- \int_{- 7}^{3} - \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

= 3 (- (- \frac{1}{2} \cdot \int_{- 7}^{3} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{2} [e^{u}]_{- 7}^{3}))

Simplify

3 (- (- \frac{1}{2} [e^{u}]_{- 7}^{3})) : \frac{3}{2} [e^{u}]_{- 7}^{3}

= \frac{3}{2} [e^{u}]_{- 7}^{3}

Compute the boundaries:

e^{3} - \frac{1}{e ^{7}}

= \frac{3}{2} (e^{3} - \frac{1}{e ^{7}})

Simplify

= \frac{3 ( e ^{10} - 1 )}{2 e ^{7}}

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What is the integral from-2 to 3 of (3e^{-2x-1}) ?
The integral from-2 to 3 of (3e^{-2x-1}) is (3(e^{10}-1))/(2e^7)