What is the integral from 0 to 1 of 5e^{-5x} ?

The integral from 0 to 1 of 5e^{-5x} is 1-1/(e^5)

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integral from 0 to 1 of 5e^{-5x}

\int_{0}^{1} 5 e^{- 5 x} d x

1 - \frac{1}{e ^{5}}

Decimal Notation

0.99326 \dots

Solution steps

\int_{0}^{1} 5 e^{- 5 x} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} e^{- 5 x} d x

Apply u-substitution

: \int_{0}^{- 5} - \frac{1}{5} e^{u} d u

= 5 \cdot \int_{0}^{- 5} - \frac{1}{5} e^{u} d u

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

= 5 (- \int_{- 5}^{0} - \frac{1}{5} e^{u} d u)

Take the constant out:

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

= 5 (- (- \frac{1}{5} \cdot \int_{- 5}^{0} e^{u} d u))

Use the common integral:

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

= 5 (- (- \frac{1}{5} [e^{u}]_{- 5}^{0}))

Simplify

5 (- (- \frac{1}{5} [e^{u}]_{- 5}^{0})) : [e^{u}]_{- 5}^{0}

= [e^{u}]_{- 5}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{5}}

= 1 - \frac{1}{e ^{5}}

\int \frac{3}{x - 1} d x

3 x^{5} - 20 x^{3}

8 x, \frac{1}{2} x, 1, 2

\int (y^{2}) d y

\frac{d y}{d x} + y^{2} = 1

What is the integral from 0 to 1 of 5e^{-5x} ?
The integral from 0 to 1 of 5e^{-5x} is 1-1/(e^5)