What is the integral from 0 to 1 of (e^{-0.2t}) ?

The integral from 0 to 1 of (e^{-0.2t}) is 0.90634…

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integral from 0 to 1 of (e^{-0.2t})

\int_{0}^{1} (e^{- 0.2 t}) d t

0.90634 \dots

Solution steps

\int_{0}^{1} e^{- 0.2 t} d t

Apply u-substitution

: \int_{0}^{- 0.2} - \frac{1}{0.2} e^{u} d u

= \int_{0}^{- 0.2} - \frac{1}{0.2} e^{u} d u

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

= - \int_{- 0.2}^{0} - \frac{1}{0.2} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.2} \cdot \int_{- 0.2}^{0} e^{u} d u)

Simplify

= - (- 5 \cdot \int_{- 0.2}^{0} e^{u} d u)

Use the common integral:

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

= - (- 5 [e^{u}]_{- 0.2}^{0})

Simplify

= 5 [e^{u}]_{- 0.2}^{0}

Compute the boundaries:

0.18126 \dots

= 5 \cdot 0.18126 \dots

Simplify

= 0.90634 \dots

\int_{0}^{1} 1 + (2 - x^{2})^{2} d x

\int arcsin (9 x) d x

y^{^{'}} + 8 y - 6 x = 0

y^{^{'}} = \frac{y}{y - x}

n = 1 \sum \infty n^{- 2.4}

What is the integral from 0 to 1 of (e^{-0.2t}) ?
The integral from 0 to 1 of (e^{-0.2t}) is 0.90634…