What is the integral from 0.5 to 1 of 3/2 (1-x^2) ?

The integral from 0.5 to 1 of 3/2 (1-x^2) is 0.3125

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integral from 0.5 to 1 of 3/2 (1-x^2)

\int_{0.5}^{1} \frac{3}{2} (1 - x^{2}) d x

0.3125

Solution steps

\int_{0.5}^{1} \frac{3}{2} (1 - x^{2}) d x

Take the constant out:

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

= \frac{3}{2} \cdot \int_{0.5}^{1} 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{3}{2} (\int_{0.5}^{1} 1 d x - \int_{0.5}^{1} x^{2} d x)

\int_{0.5}^{1} 1 d x = 0.5

\int_{0.5}^{1} x^{2} d x = 0.29166 \dots

= \frac{3}{2} (0.5 - 0.29166 \dots)

Simplify

\frac{3}{2} (0.5 - 0.29166 \dots) : 0.3125

= 0.3125

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What is the integral from 0.5 to 1 of 3/2 (1-x^2) ?
The integral from 0.5 to 1 of 3/2 (1-x^2) is 0.3125