What is the integral from 0 to 2 of 9e^{-2x} ?

The integral from 0 to 2 of 9e^{-2x} is (9(e^4-1))/(2e^4)

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integral from 0 to 2 of 9e^{-2x}

\int_{0}^{2} 9 e^{- 2 x} d x

\frac{9 ( e ^{4} - 1 )}{2 e ^{4}}

Decimal Notation

4.41757 \dots

Solution steps

\int_{0}^{2} 9 e^{- 2 x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{2} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

= 9 \cdot \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

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

= 9 (- \int_{- 4}^{0} - \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

= 9 (- (- \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u))

Use the common integral:

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

= 9 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0}))

Simplify

9 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : \frac{9}{2} [e^{u}]_{- 4}^{0}

= \frac{9}{2} [e^{u}]_{- 4}^{0}

Compute the boundaries:

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

= \frac{9}{2} (1 - \frac{1}{e ^{4}})

Simplify

= \frac{9 ( e ^{4} - 1 )}{2 e ^{4}}

n = k \sum \infty \frac{1}{2 ^{n}}

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What is the integral from 0 to 2 of 9e^{-2x} ?
The integral from 0 to 2 of 9e^{-2x} is (9(e^4-1))/(2e^4)