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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply the Power Rule

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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

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

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

\int_{1}^{5} 3 x + 1 d x

\int_{0}^{\frac{15}{16}} 1 + 16 x d x

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