What is the integral from 0 to ln(2) of 2e^{3x} ?

The integral from 0 to ln(2) of 2e^{3x} is 14/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to ln(2) of 2e^{3x}

\int_{0}^{l n (2)} 2 e^{3 x} d x

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{0}^{l n (2)} 2 e^{3 x} d x

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \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}

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

Simplify

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

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

Compute the boundaries:

7

= \frac{2}{3} \cdot 7

Simplify

= \frac{14}{3}

\int_{- 1}^{1} e^{- x^{2}} sin (x) d x

\int_{- \frac{π}{2}}^{π} 5 cos (x) d x

\int_{0}^{4} 2 x y d y

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

\int_{\frac{π}{2}}^{\frac{3 π}{2}} sin (x) d x

What is the integral from 0 to ln(2) of 2e^{3x} ?
The integral from 0 to ln(2) of 2e^{3x} is 14/3