What is the integral from 1 to infinity of 3x^{-4} ?

The integral from 1 to infinity of 3x^{-4} is 1

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integral from 1 to infinity of 3x^{-4}

\int_{1}^{\infty} 3 x^{- 4} d x

1

Solution steps

\int_{1}^{\infty} 3 x^{- 4} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} 3 x^{- 4} d x = 0 - (- 1)

= 0 - (- 1)

Simplify

= 1

\int_{0}^{1} x e^{6 x} d x

\int y - 1 d y

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = 2 e^{x}

\int 5 e^{6 x} cos (e^{6 x}) d x

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 cos (2 t) e^{- t}

What is the integral from 1 to infinity of 3x^{-4} ?
The integral from 1 to infinity of 3x^{-4} is 1