What is the integral from 1 to 2 of 6e^{6x} ?

The integral from 1 to 2 of 6e^{6x} is e^{12}-e^6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of 6e^{6x}

\int_{1}^{2} 6 e^{6 x} d x

e^{12} - e^{6}

Decimal Notation

162351.36262 \dots

Solution steps

\int_{1}^{2} 6 e^{6 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}^{2} e^{6 x} d x

Apply u-substitution

: \int_{6}^{12} e^{u} \frac{1}{6} d u

= 6 \cdot \int_{6}^{12} e^{u} \frac{1}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{6}^{12} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [e^{u}]_{6}^{12}

Simplify

6 \cdot \frac{1}{6} [e^{u}]_{6}^{12} : [e^{u}]_{6}^{12}

= [e^{u}]_{6}^{12}

Compute the boundaries:

e^{12} - e^{6}

= e^{12} - e^{6}

(2 x)^{6 x}

\int 12 cot^{4} (x) d x

\int (ln (x))^{6} d x

\frac{d}{d x} ((x^{2} + 3) 4 - x^{2})

(cos^{3} (t))^{^{'}}

What is the integral from 1 to 2 of 6e^{6x} ?
The integral from 1 to 2 of 6e^{6x} is e^{12}-e^6