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

The integral from 0 to 4 of (9e^{-2x}) is (9(e^8-1))/(2e^8)

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

\int_{0}^{4} (9 e^{- 2 x}) d x

\frac{9 ( e ^{8} - 1 )}{2 e ^{8}}

Decimal Notation

4.49849 \dots

Solution steps

\int_{0}^{4} 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}^{4} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 8} - \frac{1}{2} e^{u} d u

= 9 \cdot \int_{0}^{- 8} - \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_{- 8}^{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_{- 8}^{0} e^{u} d u))

Use the common integral:

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

= 9 (- (- \frac{1}{2} [e^{u}]_{- 8}^{0}))

Simplify

9 (- (- \frac{1}{2} [e^{u}]_{- 8}^{0})) : \frac{9}{2} [e^{u}]_{- 8}^{0}

= \frac{9}{2} [e^{u}]_{- 8}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{8}}

= \frac{9}{2} (1 - \frac{1}{e ^{8}})

Simplify

= \frac{9 ( e ^{8} - 1 )}{2 e ^{8}}

\int \frac{4}{100 - t} d t

\frac{d}{d x} ((5 + x) ln (x))

x + 1, 0.75

y = e^{x}, y = e^{- 2 x}, x = ln (5)

y = \frac{1}{2}

What is the integral from 0 to 4 of (9e^{-2x}) ?
The integral from 0 to 4 of (9e^{-2x}) is (9(e^8-1))/(2e^8)