What is the integral from 0 to 2 of 3/8 (4x-2x^2) ?

The integral from 0 to 2 of 3/8 (4x-2x^2) is 1

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integral from 0 to 2 of 3/8 (4x-2x^2)

\int_{0}^{2} \frac{3}{8} (4 x - 2 x^{2}) d x

1

Solution steps

\int_{0}^{2} \frac{3}{8} (4 x - 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}^{2} 4 x - 2 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}^{2} 4 x d x - \int_{0}^{2} 2 x^{2} d x)

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

= \frac{3}{8} (8 - \frac{16}{3})

Simplify

\frac{3}{8} (8 - \frac{16}{3}) : 1

= 1

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What is the integral from 0 to 2 of 3/8 (4x-2x^2) ?
The integral from 0 to 2 of 3/8 (4x-2x^2) is 1