What is the integral from 0 to 2 of 1/(x^{1.9)} ?

The integral from 0 to 2 of 1/(x^{1.9)} is diverges

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integral from 0 to 2 of 1/(x^{1.9)}

\int_{0}^{2} \frac{1}{x ^{1.9}} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{1.9}} d x

Apply the Power Rule

: [- \frac{1.11111 \dots}{x ^{0.9}}]_{0}^{2}

= [- \frac{1.11111 \dots}{x ^{0.9}}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

f (x) = \frac{x + 3}{x + 5}

\frac{\partial}{\partial u} (3 u^{2})

\int_{0}^{\frac{π}{2}} π cos^{2} (x) d x

\int \frac{x + 9}{x ^{2} + 4} d x

x \to \infty lim ((x!)^{2})

What is the integral from 0 to 2 of 1/(x^{1.9)} ?
The integral from 0 to 2 of 1/(x^{1.9)} is diverges