What is the integral from 0 to 3 of 2e^{-3x} ?

The integral from 0 to 3 of 2e^{-3x} is (2(e^9-1))/(3e^9)

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integral from 0 to 3 of 2e^{-3x}

\int_{0}^{3} 2 e^{- 3 x} d x

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

Decimal Notation

0.66658 \dots

Solution steps

\int_{0}^{3} 2 e^{- 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_{0}^{3} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 9} - \frac{1}{3} e^{u} d u

= 2 \cdot \int_{0}^{- 9} - \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_{- 9}^{0} - \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_{- 9}^{0} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{3} [e^{u}]_{- 9}^{0}))

Simplify

2 (- (- \frac{1}{3} [e^{u}]_{- 9}^{0})) : \frac{2}{3} [e^{u}]_{- 9}^{0}

= \frac{2}{3} [e^{u}]_{- 9}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{9}}

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

Simplify

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

\int x^{3} sin (x^{2}) d x

\frac{\partial}{\partial x} (cos^{9} (x^{4} y^{8}))

\int_{1}^{3} x + 2 d x

\frac{d}{d t} (\frac{t ^{2}}{2})

\frac{d}{d x} (e^{(x^{2} - x)})

What is the integral from 0 to 3 of 2e^{-3x} ?
The integral from 0 to 3 of 2e^{-3x} is (2(e^9-1))/(3e^9)