What is the integral from 1 to infinity of 1/(x^{1.01)} ?

The integral from 1 to infinity of 1/(x^{1.01)} is 100

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integral from 1 to infinity of 1/(x^{1.01)}

\int_{1}^{\infty} \frac{1}{x ^{1.01}} d x

100

Solution steps

\int_{1}^{\infty} \frac{1}{x ^{1.01}} d x

Compute the indefinite integral:

\int \frac{1}{x ^{1.01}} d x = - \frac{100}{x ^{0.01}} + C

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x ^{1.01}} d x = 0 - (- 100)

= 0 - (- 100)

Simplify

= 100

\frac{d}{d x} (\frac{1}{x + 10})

\frac{d}{d x} (12 ln (x))

\frac{\partial}{\partial y} (\frac{x}{x - y})

y^{^{'}} = y sin (x)

y^{^{'}} = \frac{6 x ^{2} - 1}{y}, y (0) = - 4

What is the integral from 1 to infinity of 1/(x^{1.01)} ?
The integral from 1 to infinity of 1/(x^{1.01)} is 100