What is the integral from 1 to infinity of e^{-2x} ?

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

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integral from 1 to infinity of e^{-2x}

\int_{1}^{\infty} e^{- 2 x} d x

\frac{1}{2 e ^{2}}

Solution steps

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

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

Simplify

= \frac{1}{2 e ^{2}}

\frac{\partial}{\partial x} (x^{4 y})

x y^{^{'}} = 2 y + 7 x

\frac{\partial}{\partial x} (x^{3} + 2 y^{2} + 8 x + 5 y + 5 x y)

y^{^{''}} + 9 y^{^{'}} = 40 e^{1 t}

\int 5 e^{t} d t

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