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

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

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integral from 0 to 2 of k(4x-2x^2)

\int_{0}^{2} k (4 x - 2 x^{2}) d x

\frac{8}{3} k

Solution steps

\int_{0}^{2} k (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

= k \cdot \int_{0}^{2} - 2 x^{2} + 4 x 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

= k (- \int_{0}^{2} 2 x^{2} d x + \int_{0}^{2} 4 x d x)

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

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

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

Join

- \frac{16}{3} + 8 : \frac{8}{3}

= \frac{8}{3} k

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