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

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

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

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

\frac{9}{8}

Decimal Notation

1.125

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{3}{2} - \frac{3}{8}

Simplify

\frac{3}{2} - \frac{3}{8} : \frac{9}{8}

= \frac{9}{8}

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