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

The integral from-2 to 0 of 1/(x^2) is diverges

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integral from-2 to 0 of 1/(x^2)

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

diverges

Solution steps

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

Apply the Power Rule

: [- \frac{1}{x}]_{- 2}^{0}

= [- \frac{1}{x}]_{- 2}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{2 π} (- sin^{2} (t)) d t

\int_{1}^{e} \frac{( x ^{3} + 1 )}{x} d x

\int_{0}^{\frac{π}{2}} (sin (t)) d t

\int_{0}^{1} x (x^{2}) d x

\int_{0}^{1} x^{2} 6 - 2 x^{3} d x

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