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

The integral from 1 to 3 of 2/(x^3) is 8/9

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

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

\frac{8}{9}

Decimal Notation

0.88888 \dots

Solution steps

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

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{3}

= 2 [- \frac{1}{2 x ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

= 2 \cdot \frac{4}{9}

Simplify

= \frac{8}{9}

\int \frac{x ^{n - 1}}{a + b x ^{n}} d x

\int_{0}^{1} \frac{e ^{- x}}{x} d x

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

6 x - 2 x^{2} = y, (3, 4)

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{12 e ^{4 t}}{( t ^{2} + 1 )}

What is the integral from 1 to 3 of 2/(x^3) ?
The integral from 1 to 3 of 2/(x^3) is 8/9