What is the integral from ln(1) to ln(2) of e^{3x} ?

The integral from ln(1) to ln(2) of e^{3x} is 7/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from ln(1) to ln(2) of e^{3x}

\int_{l n (1)}^{l n (2)} e^{3 x} d x

\frac{7}{3}

Decimal Notation

2.33333 \dots

Solution steps

\int_{l n (1)}^{l n (2)} e^{3 x} d x

Apply u-substitution

: \int_{0}^{3 l n (2)} e^{u} \frac{1}{3} d u

= \int_{0}^{3 l n (2)} 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

= \frac{1}{3} \cdot \int_{0}^{3 l n (2)} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{3 l n (2)}

Compute the boundaries:

7

= \frac{1}{3} \cdot 7

Simplify

= \frac{7}{3}

\frac{d}{d x} (\frac{6 x ^{5}}{5} - 7 x)

x^{2} e^{2 x}

\frac{d}{d x} (2 sec^{2} (2 x))

5 (2 x cos (3 x) - 3 x^{2} sin (3 x))

y^{^{'}} = - \frac{2 y}{x}

What is the integral from ln(1) to ln(2) of e^{3x} ?
The integral from ln(1) to ln(2) of e^{3x} is 7/3