What is the integral from 0 to 8 of 8e^{-8x} ?

The integral from 0 to 8 of 8e^{-8x} is 1-1/(e^{64)}

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integral from 0 to 8 of 8e^{-8x}

\int_{0}^{8} 8 e^{- 8 x} d x

1 - \frac{1}{e ^{64}}

Decimal Notation

1

Solution steps

\int_{0}^{8} 8 e^{- 8 x} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{8} e^{- 8 x} d x

Apply u-substitution

: \int_{0}^{- 64} - \frac{1}{8} e^{u} d u

= 8 \cdot \int_{0}^{- 64} - \frac{1}{8} e^{u} d u

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

= 8 (- \int_{- 64}^{0} - \frac{1}{8} e^{u} d u)

Take the constant out:

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

= 8 (- (- \frac{1}{8} \cdot \int_{- 64}^{0} e^{u} d u))

Use the common integral:

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

= 8 (- (- \frac{1}{8} [e^{u}]_{- 64}^{0}))

Simplify

8 (- (- \frac{1}{8} [e^{u}]_{- 64}^{0})) : [e^{u}]_{- 64}^{0}

= [e^{u}]_{- 64}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{64}}

= 1 - \frac{1}{e ^{64}}

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What is the integral from 0 to 8 of 8e^{-8x} ?
The integral from 0 to 8 of 8e^{-8x} is 1-1/(e^{64)}