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

The integral from 2 to 3 of 1/(x^2)-x is -7/3

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

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

- \frac{7}{3}

Decimal Notation

- 2.33333 \dots

Solution steps

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

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

\int_{2}^{3} x d x = \frac{5}{2}

= \frac{1}{6} - \frac{5}{2}

Simplify

\frac{1}{6} - \frac{5}{2} : - \frac{7}{3}

= - \frac{7}{3}

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What is the integral from 2 to 3 of 1/(x^2)-x ?
The integral from 2 to 3 of 1/(x^2)-x is -7/3