What is the integral from 0 to 5 of 6e^{2x} ?

The integral from 0 to 5 of 6e^{2x} is 3(e^{10}-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of 6e^{2x}

\int_{0}^{5} 6 e^{2 x} d x

3 (e^{10} - 1)

Decimal Notation

66076.39738 \dots

Solution steps

\int_{0}^{5} 6 e^{2 x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{5} e^{2 x} d x

Apply u-substitution

: \int_{0}^{10} e^{u} \frac{1}{2} d u

= 6 \cdot \int_{0}^{10} e^{u} \frac{1}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{10} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{10}

Simplify

6 \cdot \frac{1}{2} [e^{u}]_{0}^{10} : 3 [e^{u}]_{0}^{10}

= 3 [e^{u}]_{0}^{10}

Compute the boundaries:

e^{10} - 1

= 3 (e^{10} - 1)

\int_{- 1}^{0} x^{2} x + 1 d x

\int_{- 1}^{1} t^{3} (2 + t^{4})^{3} d t

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 3} d x

\int_{- 1}^{4} x^{2} d x

What is the integral from 0 to 5 of 6e^{2x} ?
The integral from 0 to 5 of 6e^{2x} is 3(e^{10}-1)