What is the integral from 3 to infinity of e^{-3x} ?

The integral from 3 to infinity of e^{-3x} is 1/(3e^9)

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integral from 3 to infinity of e^{-3x}

\int_{3}^{\infty} e^{- 3 x} d x

\frac{1}{3 e ^{9}}

Decimal Notation

0.00004 \dots

Solution steps

\int_{3}^{\infty} e^{- 3 x} d x

Compute the indefinite integral:

\int e^{- 3 x} d x = - \frac{1}{3} e^{- 3 x} + C

Compute the boundaries:

\int_{3}^{\infty} e^{- 3 x} d x = 0 - (- \frac{1}{3 e ^{9}})

= 0 - (- \frac{1}{3 e ^{9}})

Simplify

= \frac{1}{3 e ^{9}}

y^{^{''}} - 2 y^{^{'}} + 82 y = 0

x \to 0 + lim ((1 + x)^{x})

x \to 0 + lim (tan (5 x) ln (8 x))

x \to \frac{1}{4} lim (tan (π x))

What is the integral from 3 to infinity of e^{-3x} ?
The integral from 3 to infinity of e^{-3x} is 1/(3e^9)