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

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

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

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

1

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

= 1

f (x) = (2 x^{3} + 5)^{10}

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\int x sin (π) x d x

\frac{\partial}{\partial x} (4 x^{2} \cdot 3 x^{3} y^{2})

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