What is the integral from 0.5 to 1 of 3/8 (2-x)^2 ?

The integral from 0.5 to 1 of 3/8 (2-x)^2 is 0.296875

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integral from 0.5 to 1 of 3/8 (2-x)^2

\int_{0.5}^{1} \frac{3}{8} (2 - x)^{2} d x

0.296875

Solution steps

\int_{0.5}^{1} \frac{3}{8} (2 - x)^{2} d x

Take the constant out:

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

= \frac{3}{8} \cdot \int_{0.5}^{1} (2 - x)^{2} d x

Expand

(2 - x)^{2} : 4 - 4 x + x^{2}

= \frac{3}{8} \cdot \int_{0.5}^{1} 4 - 4 x + 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}{8} (\int_{0.5}^{1} 4 d x - \int_{0.5}^{1} 4 x d x + \int_{0.5}^{1} x^{2} d x)

\int_{0.5}^{1} 4 d x = 2

\int_{0.5}^{1} 4 x d x = 1.5

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

= \frac{3}{8} (2 - 1.5 + 0.29166 \dots)

Simplify

\frac{3}{8} (2 - 1.5 + 0.29166 \dots) : 0.296875

= 0.296875

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What is the integral from 0.5 to 1 of 3/8 (2-x)^2 ?
The integral from 0.5 to 1 of 3/8 (2-x)^2 is 0.296875