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

The integral from 0 to ln(3) of 2e^{3x} is 52/3

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

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

\frac{52}{3}

Solution steps

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

Apply u-substitution

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

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

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

Simplify

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

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

Compute the boundaries:

26

= \frac{2}{3} \cdot 26

Simplify

= \frac{52}{3}

\int (x^{3} + e x) x^{2} + e d x

- 4 x sin (2 x^{2})

f (x) = - 8 (4 x + 3)^{- 3}, at x = 3

\frac{d}{d x} (cot (x) + csc (x))

What is the integral from 0 to ln(3) of 2e^{3x} ?
The integral from 0 to ln(3) of 2e^{3x} is 52/3