What is the integral from 0 to 4 of e^{-3x} ?

The integral from 0 to 4 of e^{-3x} is (e^{12}-1)/(3e^{12)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of e^{-3x}

\int_{0}^{4} e^{- 3 x} d x

\frac{e ^{12} - 1}{3 e ^{12}}

Decimal

0.33333 \dots

Solution steps

\int_{0}^{4} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

= \int_{0}^{- 12} - \frac{1}{3} e^{u} d u

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

= - \int_{- 12}^{0} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 12}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 12}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 12}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{12}}

= \frac{1}{3} (1 - \frac{1}{e ^{12}})

Simplify

= \frac{e ^{12} - 1}{3 e ^{12}}

\int_{\frac{π}{2}}^{π} (sin (x)) d x

\int_{0}^{2} \frac{x}{1 + 4 x ^{2}} d x

\int_{0}^{1} 5 x^{4} (6 - x^{5}) d x

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

\int_{1}^{2} \frac{1}{x} + x d x

What is the integral from 0 to 4 of e^{-3x} ?
The integral from 0 to 4 of e^{-3x} is (e^{12}-1)/(3e^{12)}