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

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

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

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

\frac{9}{4}

Decimal

2.25

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{3}{8}

= 6 \cdot \frac{3}{8}

Simplify

= \frac{9}{4}

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