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

The integral from 0 to infinity of e^{(-2x)} is 1/2

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

\int_{0}^{\infty} e^{(- 2 x)} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{\infty} e^{- 2 x} d x

Compute the indefinite integral:

\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x} + C

Compute the boundaries:

\int_{0}^{\infty} e^{(- 2 x)} d x = 0 - (- \frac{1}{2})

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

4^{(x^{9})}

\frac{1}{( 2 + x ) ^{3}}

x \to - \infty lim (3^{- x})

(\frac{11}{6}, - \frac{1}{2}), (\frac{1}{3}, \frac{1}{3})

(1) (3.5)

What is the integral from 0 to infinity of e^{(-2x)} ?
The integral from 0 to infinity of e^{(-2x)} is 1/2