What is the integral from 1 to 4 of e^{bx-b} ?

The integral from 1 to 4 of e^{bx-b} is (e^{-b}(e^{4b}-e^b))/b

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of e^{bx-b}

\int_{1}^{4} e^{b x - b} d x

\frac{e ^{- b} ( e ^{4 b} - e ^{b} )}{b}

Solution steps

\int_{1}^{4} e^{b x - b} d x

e^{b x - b} = e^{- b} e^{b x}

= \int_{1}^{4} e^{- b} e^{b x} d x

Take the constant out:

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

= e^{- b} \cdot \int_{1}^{4} e^{b x} d x

Apply u-substitution

: \int_{b}^{b \cdot 4} \frac{e ^{u}}{b} d u

= e^{- b} \cdot \int_{b}^{b \cdot 4} \frac{e ^{u}}{b} d u

Take the constant out:

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

= e^{- b} \frac{1}{b} \cdot \int_{b}^{b \cdot 4} e^{u} d u

Use the common integral:

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

= e^{- b} \frac{1}{b} [e^{u}]_{b}^{b \cdot 4}

Compute the boundaries:

e^{b \cdot 4} - e^{b}

= e^{- b} \frac{1}{b} (e^{b \cdot 4} - e^{b})

Simplify

= \frac{e ^{- b} ( e ^{4 b} - e ^{b} )}{b}

\int_{0}^{1} s + t d s

\int_{1}^{8} 3 x^{2} d x

\int_{0}^{T} (x^{4} - 8 x + 7) d x

\int_{0}^{1} \frac{1}{e ^{x} - e ^{- x}} d x

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

What is the integral from 1 to 4 of e^{bx-b} ?
The integral from 1 to 4 of e^{bx-b} is (e^{-b}(e^{4b}-e^b))/b