What is the integral from 0 to 3 of 1/5 e^{-x/5} ?

The integral from 0 to 3 of 1/5 e^{-x/5} is 1-1/(e^{3/5)}

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integral from 0 to 3 of 1/5 e^{-x/5}

\int_{0}^{3} \frac{1}{5} e^{- \frac{x}{5}} d x

1 - \frac{1}{e ^{\frac{3}{5}}}

Decimal

0.45118 \dots

Solution steps

\int_{0}^{3} \frac{1}{5} e^{- \frac{x}{5}} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{3} e^{- \frac{x}{5}} d x

Apply u-substitution

: \int_{0}^{- \frac{3}{5}} - 5 e^{u} d u

= \frac{1}{5} \cdot \int_{0}^{- \frac{3}{5}} - 5 e^{u} d u

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

= \frac{1}{5} (- \int_{- \frac{3}{5}}^{0} - 5 e^{u} d u)

Take the constant out:

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

= \frac{1}{5} (- (- 5 \cdot \int_{- \frac{3}{5}}^{0} e^{u} d u))

Use the common integral:

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

= \frac{1}{5} (- (- 5 [e^{u}]_{- \frac{3}{5}}^{0}))

Simplify

\frac{1}{5} (- (- 5 [e^{u}]_{- \frac{3}{5}}^{0})) : [e^{u}]_{- \frac{3}{5}}^{0}

= [e^{u}]_{- \frac{3}{5}}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{\frac{3}{5}}}

= 1 - \frac{1}{e ^{\frac{3}{5}}}

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What is the integral from 0 to 3 of 1/5 e^{-x/5} ?
The integral from 0 to 3 of 1/5 e^{-x/5} is 1-1/(e^{3/5)}