What is the integral from 0 to 1 of 8e^{-4x} ?

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

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integral from 0 to 1 of 8e^{-4x}

\int_{0}^{1} 8 e^{- 4 x} d x

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

Decimal Notation

1.96336 \dots

Solution steps

\int_{0}^{1} 8 e^{- 4 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}^{1} e^{- 4 x} d x

Apply u-substitution

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

= 8 \cdot \int_{0}^{- 4} - \frac{1}{4} e^{u} d u

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

= 8 (- \int_{- 4}^{0} - \frac{1}{4} 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}{4} \cdot \int_{- 4}^{0} e^{u} d u))

Use the common integral:

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

= 8 (- (- \frac{1}{4} [e^{u}]_{- 4}^{0}))

Simplify

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

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

Compute the boundaries:

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

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

(0.2) (3.5)

x \to 5 lim (x^{2} - 16)

\int - \frac{y}{x} d x

f (x) = (8 x^{2} - 14 x) e^{x}

\frac{3}{( s + 1 )}

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