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

The integral from 2 to infinity of e^{-2x} is 1/(2e^4)

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

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

\frac{1}{2 e ^{4}}

Decimal Notation

0.00915 \dots

Solution steps

\int_{2}^{\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_{2}^{\infty} e^{- 2 x} d x = 0 - (- \frac{1}{2 e ^{4}})

= 0 - (- \frac{1}{2 e ^{4}})

Simplify

= \frac{1}{2 e ^{4}}

f (x) = x cos (x^{2})

\frac{\partial}{\partial y} (x^{3} - 6 x y + y^{3})

x \to 5 lim (2 x)

f (x) = 4 x^{2} + 12 x + 3, at x = - 2

\int 5 x^{4} - 8 x^{5} d x

What is the integral from 2 to infinity of e^{-2x} ?
The integral from 2 to infinity of e^{-2x} is 1/(2e^4)