What is the integral from 1 to 3 of (x^2-1/(4x^2)) ?

The integral from 1 to 3 of (x^2-1/(4x^2)) is 17/2

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integral from 1 to 3 of (x^2-1/(4x^2))

\int_{1}^{3} (x^{2} - \frac{1}{4 x ^{2}}) d x

\frac{17}{2}

Decimal

8.5

Solution steps

\int_{1}^{3} x^{2} - \frac{1}{4 x ^{2}} 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}^{3} x^{2} d x - \int_{1}^{3} \frac{1}{4 x ^{2}} d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} \frac{1}{4 x ^{2}} d x = \frac{1}{6}

= \frac{26}{3} - \frac{1}{6}

Simplify

\frac{26}{3} - \frac{1}{6} : \frac{17}{2}

= \frac{17}{2}

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What is the integral from 1 to 3 of (x^2-1/(4x^2)) ?
The integral from 1 to 3 of (x^2-1/(4x^2)) is 17/2