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 1/3

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Take the constant out:

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

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

Expand

\frac{x ^{2} - 1}{x ^{2}} : 1 - \frac{1}{x ^{2}}

= \frac{1}{4} \cdot \int_{1}^{3} 1 - \frac{1}{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

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

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

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

= \frac{1}{4} (2 - \frac{2}{3})

Simplify

\frac{1}{4} (2 - \frac{2}{3}) : \frac{1}{3}

= \frac{1}{3}

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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 1/3