What is the integral from 0 to infinity of e^{-1/5} ?

The integral from 0 to infinity of e^{-1/5} is diverges

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integral from 0 to infinity of e^{-1/5}

\int_{0}^{\infty} e^{- \frac{1}{5}}

diverges

Solution steps

\int_{0}^{\infty} e^{- \frac{1}{5}} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\infty} e^{- \frac{1}{5}} d x =

diverges

= diverges

\int_{- 1}^{1} 4 (1 - x^{2}) - (1 - x^{2}) d x

\int_{1}^{2} x^{2} - x + 3 d x

\int_{6}^{100} 0.5 x^{- 1.5} d x

\int_{1}^{2} x^{2} - x d x

\int_{0}^{10} 9 y^{\frac{1}{3}} d y

What is the integral from 0 to infinity of e^{-1/5} ?
The integral from 0 to infinity of e^{-1/5} is diverges