What is the integral from 0 to 60 of 25e^{-0.03x} ?

The integral from 0 to 60 of 25e^{-0.03x} is 695.58425…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 60 of 25e^{-0.03x}

\int_{0}^{60} 25 e^{- 0.03 x} d x

695.58425 \dots

Solution steps

\int_{0}^{60} 25 e^{- 0.03 x} d x

Take the constant out:

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

= 25 \cdot \int_{0}^{60} e^{- 0.03 x} d x

Apply u-substitution

: \int_{0}^{- 1.8} - \frac{1}{0.03} e^{u} d u

= 25 \cdot \int_{0}^{- 1.8} - \frac{1}{0.03} e^{u} d u

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

= 25 (- \int_{- 1.8}^{0} - \frac{1}{0.03} 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}{0.03} \cdot \int_{- 1.8}^{0} e^{u} d u))

Simplify

= 25 (- (- 33.33333 \dots \cdot \int_{- 1.8}^{0} e^{u} d u))

Use the common integral:

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

= 25 (- (- 33.33333 \dots [e^{u}]_{- 1.8}^{0}))

Simplify

= 833.33333 \dots [e^{u}]_{- 1.8}^{0}

Compute the boundaries:

0.83470 \dots

= 833.33333 \dots \cdot 0.83470 \dots

Simplify

= 695.58425 \dots

\int \frac{9 t - 4}{t + 1} d t

\frac{d}{d x} (x y z + \frac{5}{4} z^{4} + e^{x} cos (y))

x \to 0 lim (x sin (\frac{3}{x}))

\int sin^{2} (o) d o

\frac{( s + 3 ) ^{2}}{s ( s ^{2} + 10 )}

What is the integral from 0 to 60 of 25e^{-0.03x} ?
The integral from 0 to 60 of 25e^{-0.03x} is 695.58425…