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

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

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

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

ln (2) - \frac{3}{2}

Decimal

- 0.80685 \dots

Solution steps

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

Expand

\frac{1 - ( \frac{x}{2} ) ^{2}}{x} : \frac{1}{x} - \frac{x}{4}

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

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

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

= ln (2) - \frac{3}{2}

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