What is the integral from 1 to 4 of 6e^{sqrt(x)} ?

The integral from 1 to 4 of 6e^{sqrt(x)} is 12e^2

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 6e^{sqrt(x)}

\int_{1}^{4} 6 e^{x} d x

12 e^{2}

Decimal Notation

88.66867 \dots

Solution steps

\int_{1}^{4} 6 e^{x} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{4} e^{x} d x

Apply u-substitution

: \int_{1}^{2} e^{u} \cdot 2 u d u

= 6 \cdot \int_{1}^{2} e^{u} \cdot 2 u d u

Take the constant out:

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

= 6 \cdot 2 \cdot \int_{1}^{2} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{1}^{2}

= 6 \cdot 2 [e^{u} u - \int e^{u} d u]_{1}^{2}

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

= 6 \cdot 2 [e^{u} u - e^{u}]_{1}^{2}

Simplify

= 12 [e^{u} u - e^{u}]_{1}^{2}

Compute the boundaries:

e^{2}

= 12 e^{2}

f (x, y) = (x)^{2} + (y)^{4}

x \to 3 + lim (\frac{1}{∣ x - 3 ∣})

y \to 0 lim (\frac{y ^{2}}{y ^{4}})

f (x) = e^{x} - x^{7}

x \to 1 lim (\frac{x ^{2} - 9}{x - 3})

What is the integral from 1 to 4 of 6e^{sqrt(x)} ?
The integral from 1 to 4 of 6e^{sqrt(x)} is 12e^2