What is the integral from-2 to-1 of 6x^{-3} ?

The integral from-2 to-1 of 6x^{-3} is -9/4

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integral from-2 to-1 of 6x^{-3}

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

- \frac{9}{4}

Decimal Notation

- 2.25

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

= 6 [- \frac{1}{2 x ^{2}}]_{- 2}^{- 1}

Compute the boundaries:

- \frac{3}{8}

= 6 (- \frac{3}{8})

Simplify

= - \frac{9}{4}

\frac{d}{d x} (\frac{x ^{2} + x}{x + 1})

\frac{d}{d x} (\frac{x ^{2} + 2}{x ^{4} - 3 x ^{2} + 1})

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 4

\int x^{2} + 1 x d x

x \to 0 lim (sin (x) \cdot ln (4 x))

What is the integral from-2 to-1 of 6x^{-3} ?
The integral from-2 to-1 of 6x^{-3} is -9/4