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

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

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

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

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

Decimal Notation

0.13533 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\frac{\partial}{\partial x} ((sin (x - y)) (- 1.1))

y = \frac{4}{3} π r^{3}

ln (4)

\int \frac{x ^{2} - 1}{x ^{3} + x ^{2} + x} d x

f (x) = x^{2} - 7 x + 5, at x = 4

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