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

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

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

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

\frac{3}{2}

Decimal Notation

1.5

Solution steps

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

Apply u-substitution

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

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

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

Use the common integral:

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

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

Compute the boundaries:

3

= \frac{1}{2} \cdot 3

Simplify

= \frac{3}{2}

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

y^{^{'}} = x^{3} (1 - y), y (0) = 3

\int \frac{x - 1}{x + 1} d x

e^{- 1}

t \to 0 lim (e^{- 7 t})

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