What is the integral from 0 to 1 of 4e^{6sqrt(x)} ?

The integral from 0 to 1 of 4e^{6sqrt(x)} is (2(5e^6+1))/9

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 4e^{6sqrt(x)}

\int_{0}^{1} 4 e^{6 x} d x

\frac{2 ( 5 e ^{6} + 1 )}{9}

Decimal Notation

448.47643 \dots

Solution steps

\int_{0}^{1} 4 e^{6 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} e^{6 x} d x

Apply u-substitution

: \int_{0}^{6} \frac{e ^{u} u}{18} d u

= 4 \cdot \int_{0}^{6} \frac{e ^{u} u}{18} d u

Take the constant out:

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

= 4 \cdot \frac{1}{18} \cdot \int_{0}^{6} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{6}

= 4 \cdot \frac{1}{18} [e^{u} u - \int e^{u} d u]_{0}^{6}

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

= 4 \cdot \frac{1}{18} [e^{u} u - e^{u}]_{0}^{6}

Simplify

4 \cdot \frac{1}{18} [e^{u} u - e^{u}]_{0}^{6} : \frac{2}{9} [e^{u} u - e^{u}]_{0}^{6}

= \frac{2}{9} [e^{u} u - e^{u}]_{0}^{6}

Compute the boundaries:

5 e^{6} + 1

= \frac{2}{9} (5 e^{6} + 1)

Simplify

= \frac{2 ( 5 e ^{6} + 1 )}{9}

n = 0 \sum \infty \frac{3 ^{n}}{n !}

\frac{d}{d x} ((x) cos (x) sin (x))

\frac{\partial}{\partial x} (e^{2 (2 x + y)})

ln (x^{2} - 1)

n = 0 \sum \infty n x^{n - 1}

What is the integral from 0 to 1 of 4e^{6sqrt(x)} ?
The integral from 0 to 1 of 4e^{6sqrt(x)} is (2(5e^6+1))/9