What is the integral from 0 to 5 of 9e^{3x} ?

The integral from 0 to 5 of 9e^{3x} is 3(e^{15}-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 9e^{3x}

\int_{0}^{5} 9 e^{3 x} d x

3 (e^{15} - 1)

Decimal Notation

9807049.11741 \dots

Solution steps

\int_{0}^{5} 9 e^{3 x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{5} e^{3 x} d x

Apply u-substitution

: \int_{0}^{15} e^{u} \frac{1}{3} d u

= 9 \cdot \int_{0}^{15} 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

= 9 \cdot \frac{1}{3} \cdot \int_{0}^{15} e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} [e^{u}]_{0}^{15}

Simplify

9 \cdot \frac{1}{3} [e^{u}]_{0}^{15} : 3 [e^{u}]_{0}^{15}

= 3 [e^{u}]_{0}^{15}

Compute the boundaries:

e^{15} - 1

= 3 (e^{15} - 1)

\int_{0}^{4} \frac{p}{9 + p ^{2}} d p

\int_{0}^{4} 2 x^{2} + 5 d x

\int_{- 1}^{1} ∣ x + 2 ∣ d x

\int_{- 2021}^{2021} x^{5} d x

What is the integral from 0 to 5 of 9e^{3x} ?
The integral from 0 to 5 of 9e^{3x} is 3(e^{15}-1)