What is the integral from 0 to 6 of sqrt(4-2x) ?

The integral from 0 to 6 of sqrt(4-2x) is 8/3+i(16sqrt(2))/3

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integral from 0 to 6 of sqrt(4-2x)

\int_{0}^{6} 4 - 2 x d x

\frac{8}{3} + i \frac{16 2}{3}

Solution steps

\int_{0}^{6} 4 - 2 x d x

Apply u-substitution

: \int_{4}^{- 8} - \frac{u}{2} d u

= \int_{4}^{- 8} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 8}^{4} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 8}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{- 8}^{4}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 8}^{4})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{- 8}^{4}

Compute the boundaries:

\frac{16}{3} + i \frac{32 2}{3}

= \frac{1}{2} (\frac{16}{3} + i \frac{32 2}{3})

Simplify

= \frac{8}{3} + i \frac{16 2}{3}

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What is the integral from 0 to 6 of sqrt(4-2x) ?
The integral from 0 to 6 of sqrt(4-2x) is 8/3+i(16sqrt(2))/3