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

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

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

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

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

Decimal Notation

0.01831 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int 2 b x^{3} d x

\frac{d}{d x} ((5 x^{2} - 7) (3 x^{2} - 2 x + 1))

y = \frac{x}{4 - x}, (3, 3)

\frac{d}{d x} ((x - 2)^{2} - ln (x))

- \frac{1}{4 x ^{\frac{3}{2}}}

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