What is the integral from 0 to 1 of 12e^{3x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of 12e^{3x}

\int_{0}^{1} 12 e^{3 x} d x

4 (e^{3} - 1)

Decimal

76.34214 \dots

Solution steps

\int_{0}^{1} 12 e^{3 x} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{1} e^{3 x} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

12 \cdot \frac{1}{3} [e^{u}]_{0}^{3} : 4 [e^{u}]_{0}^{3}

= 4 [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

= 4 (e^{3} - 1)

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