What is the integral from 0 to 1 of (e^{-0.1t}) ?

The integral from 0 to 1 of (e^{-0.1t}) is 0.95162…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (e^{-0.1t})

\int_{0}^{1} (e^{- 0.1 t}) d t

0.95162 \dots

Solution steps

\int_{0}^{1} e^{- 0.1 t} d t

Apply u-substitution

: \int_{0}^{- 0.1} - \frac{1}{0.1} e^{u} d u

= \int_{0}^{- 0.1} - \frac{1}{0.1} e^{u} d u

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

= - \int_{- 0.1}^{0} - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.1} \cdot \int_{- 0.1}^{0} e^{u} d u)

Simplify

= - (- 10 \cdot \int_{- 0.1}^{0} e^{u} d u)

Use the common integral:

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

= - (- 10 [e^{u}]_{- 0.1}^{0})

Simplify

= 10 [e^{u}]_{- 0.1}^{0}

Compute the boundaries:

0.09516 \dots

= 10 \cdot 0.09516 \dots

Simplify

= 0.95162 \dots

\int_{0}^{π} 2 sin (n x) d x

f (x) = 14 x

\frac{d}{d x} (2 arctan (e^{x}))

\frac{d}{d x} (8 (x + 2)^{\frac{3}{2}})

\frac{d}{d x} (- 2 x^{4} + 3 x^{3} - x^{2})

What is the integral from 0 to 1 of (e^{-0.1t}) ?
The integral from 0 to 1 of (e^{-0.1t}) is 0.95162…