What is the integral from 0.5 to 2 of 0.09375(4-x^2) ?

The integral from 0.5 to 2 of 0.09375(4-x^2) is 0.31640625

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integral from 0.5 to 2 of 0.09375(4-x^2)

\int_{0.5}^{2} 0.09375 (4 - x^{2}) d x

0.31640625

Solution steps

\int_{0.5}^{2} 0.09375 (4 - x^{2}) d x

Take the constant out:

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

= 0.09375 \cdot \int_{0.5}^{2} 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

= 0.09375 (\int_{0.5}^{2} 4 d x - \int_{0.5}^{2} x^{2} d x)

\int_{0.5}^{2} 4 d x = 6

\int_{0.5}^{2} x^{2} d x = 2.625

= 0.09375 (6 - 2.625)

Simplify

= 0.31640625

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What is the integral from 0.5 to 2 of 0.09375(4-x^2) ?
The integral from 0.5 to 2 of 0.09375(4-x^2) is 0.31640625