What is the integral from 1 to infinity of 6e^{-3x} ?

The integral from 1 to infinity of 6e^{-3x} is 2/(e^3)

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integral from 1 to infinity of 6e^{-3x}

\int_{1}^{\infty} 6 e^{- 3 x} d x

\frac{2}{e ^{3}}

Decimal

0.09957 \dots

Solution steps

\int_{1}^{\infty} 6 e^{- 3 x} d x

Compute the indefinite integral:

\int 6 e^{- 3 x} d x = - 2 e^{- 3 x} + C

Compute the boundaries:

\int_{1}^{\infty} 6 e^{- 3 x} d x = 0 - (- \frac{2}{e ^{3}})

= 0 - (- \frac{2}{e ^{3}})

Simplify

= \frac{2}{e ^{3}}

u^{\frac{1}{2}}

\int_{0}^{4} 3 e^{- 2 x} d x

\frac{\partial}{\partial x} (x^{1} x^{2} + 10 x^{1})

y^{3} - (18 x + 8) + 3 x y^{2} (\frac{d y}{d x}) = 0

x \to 1 lim (6)

What is the integral from 1 to infinity of 6e^{-3x} ?
The integral from 1 to infinity of 6e^{-3x} is 2/(e^3)