What is the integral from 2 to infinity of e^{-5x} ?

The integral from 2 to infinity of e^{-5x} is 1/(5e^{10)}

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integral from 2 to infinity of e^{-5x}

\int_{2}^{\infty} e^{- 5 x} d x

\frac{1}{5 e ^{10}}

Decimal

9.07999 E - 6

Solution steps

\int_{2}^{\infty} e^{- 5 x} d x

Compute the indefinite integral:

\int e^{- 5 x} d x = - \frac{1}{5} e^{- 5 x} + C

Compute the boundaries:

\int_{2}^{\infty} e^{- 5 x} d x = 0 - (- \frac{1}{5 e ^{10}})

= 0 - (- \frac{1}{5 e ^{10}})

Simplify

= \frac{1}{5 e ^{10}}

\int 7 ln (x^{2} - x + 2) d x

\frac{d}{d x} (2 x^{2} + cos^{2} (x))

\frac{d}{d x} (cos (\frac{3 x ^{2}}{x + 2}))

x \to 2 lim ([7 f (x) + g (x)])

\frac{d y}{d x}, x y = 56

What is the integral from 2 to infinity of e^{-5x} ?
The integral from 2 to infinity of e^{-5x} is 1/(5e^{10)}