What is the integral from 1 to 8 of 2/(x^3) ?

The integral from 1 to 8 of 2/(x^3) is 63/64

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integral from 1 to 8 of 2/(x^3)

\int_{1}^{8} \frac{2}{x ^{3}} d x

\frac{63}{64}

Solution steps

\int_{1}^{8} \frac{2}{x ^{3}} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{8} \frac{1}{x ^{3}} d x

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{8}

= 2 [- \frac{1}{2 x ^{2}}]_{1}^{8}

Compute the boundaries:

\frac{63}{128}

= 2 \cdot \frac{63}{128}

Simplify

= \frac{63}{64}

x \to 0 + lim ((- ln (x))^{x})

\frac{d}{d x} (x^{2} x^{4} - 12)

\frac{d y}{d x} = \frac{( 3 x ^{2} )}{y ^{2}}

\frac{d}{d x} (x - sin (2 x))

What is the integral from 1 to 8 of 2/(x^3) ?
The integral from 1 to 8 of 2/(x^3) is 63/64