What is the integral from 0 to 0.2 of 25e^{-5y} ?

The integral from 0 to 0.2 of 25e^{-5y} is 5(1-1/e)

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integral from 0 to 0.2 of 25e^{-5y}

\int_{0}^{0.2} 25 e^{- 5 y} d y

5 (1 - \frac{1}{e})

Decimal Notation

3.16060 \dots

Solution steps

\int_{0}^{0.2} 25 e^{- 5 y} d y

Take the constant out:

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

= 25 \cdot \int_{0}^{0.2} e^{- 5 y} d y

Apply u-substitution

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

= 25 \cdot \int_{0}^{- 1} - \frac{1}{5} e^{u} d u

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

= 25 (- \int_{- 1}^{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

= 25 (- (- \frac{1}{5} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

= 25 (- (- \frac{1}{5} [e^{u}]_{- 1}^{0}))

Simplify

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

= 5 [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 5 (1 - \frac{1}{e})

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What is the integral from 0 to 0.2 of 25e^{-5y} ?
The integral from 0 to 0.2 of 25e^{-5y} is 5(1-1/e)