What is the integral from-infinity to-8 of x^{-4/3} ?

The integral from-infinity to-8 of x^{-4/3} is diverges

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integral from-infinity to-8 of x^{-4/3}

\int_{- \infty}^{- 8} x^{- \frac{4}{3}} d x

diverges

Solution steps

\int_{- \infty}^{- 8} x^{- \frac{4}{3}} d x

Compute the indefinite integral:

\int x^{- \frac{4}{3}} d x = - \frac{3}{x ^{\frac{1}{3}}} + C

Compute the boundaries:

\int_{- \infty}^{- 8} x^{- \frac{4}{3}} d x =

diverges

= diverges

\frac{d x}{d t} = 6 - \frac{6 x}{15}

x \to \infty lim (2 - \frac{1}{x})

\int \frac{4 x + 1}{x ^{2}} d x

y^{^{'}} = 3 + t - y, y (0) = 1

\int_{1}^{3} (x^{2} - 5 x + 3) d x

What is the integral from-infinity to-8 of x^{-4/3} ?
The integral from-infinity to-8 of x^{-4/3} is diverges