What is the integral from 0 to 10 of e^{-0.04t} ?

The integral from 0 to 10 of e^{-0.04t} is 8.24199…

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integral from 0 to 10 of e^{-0.04t}

\int_{0}^{10} e^{- 0.04 t} d t

8.24199 \dots

Solution steps

\int_{0}^{10} e^{- 0.04 t} d t

Apply u-substitution

: \int_{0}^{- 0.4} - \frac{1}{0.04} e^{u} d u

= \int_{0}^{- 0.4} - \frac{1}{0.04} e^{u} d u

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

= - \int_{- 0.4}^{0} - \frac{1}{0.04} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.04} \cdot \int_{- 0.4}^{0} e^{u} d u)

Simplify

= - (- 25 \cdot \int_{- 0.4}^{0} e^{u} d u)

Use the common integral:

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

= - (- 25 [e^{u}]_{- 0.4}^{0})

Simplify

= 25 [e^{u}]_{- 0.4}^{0}

Compute the boundaries:

0.32967 \dots

= 25 \cdot 0.32967 \dots

Simplify

= 8.24199 \dots

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What is the integral from 0 to 10 of e^{-0.04t} ?
The integral from 0 to 10 of e^{-0.04t} is 8.24199…