What is the integral from 0 to 1 of xsqrt(4-5x) ?

The integral from 0 to 1 of xsqrt(4-5x) is 128/375+i 46/375

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integral from 0 to 1 of xsqrt(4-5x)

\int_{0}^{1} x 4 - 5 x d x

\frac{128}{375} + i \frac{46}{375}

Solution steps

\int_{0}^{1} x 4 - 5 x d x

Apply u-substitution

= \int_{4}^{- 1} \frac{u ( u - 4 )}{25} d u

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

= - \int_{- 1}^{4} \frac{u ( u - 4 )}{25} d u

Take the constant out:

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

= - \frac{1}{25} \cdot \int_{- 1}^{4} u (u - 4) d u

Expand

u (u - 4) : u^{\frac{3}{2}} - 4 u

= - \frac{1}{25} \cdot \int_{- 1}^{4} u^{\frac{3}{2}} - 4 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{25} (\int_{- 1}^{4} u^{\frac{3}{2}} d u - \int_{- 1}^{4} 4 u d u)

\int_{- 1}^{4} u^{\frac{3}{2}} d u = \frac{64}{5} - i \frac{2}{5}

\int_{- 1}^{4} 4 u d u = \frac{64}{3} + i \frac{8}{3}

= - \frac{1}{25} (\frac{64}{5} - i \frac{2}{5} - (\frac{64}{3} + i \frac{8}{3}))

Simplify

- \frac{1}{25} (\frac{64}{5} - i \frac{2}{5} - (\frac{64}{3} + i \frac{8}{3})) : \frac{128}{375} + i \frac{46}{375}

= \frac{128}{375} + i \frac{46}{375}

What is the integral from 0 to 1 of xsqrt(4-5x) ?
The integral from 0 to 1 of xsqrt(4-5x) is 128/375+i 46/375