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

The integral from 0 to 1 of xsqrt(4+5x) is 506/375

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

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

\frac{506}{375}

Decimal

1.34933 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{9} \frac{u ( u - 4 )}{25} d u

= \int_{4}^{9} \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_{4}^{9} u (u - 4) d u

Expand

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

= \frac{1}{25} \cdot \int_{4}^{9} 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_{4}^{9} u^{\frac{3}{2}} d u - \int_{4}^{9} 4 u d u)

\int_{4}^{9} u^{\frac{3}{2}} d u = \frac{422}{5}

\int_{4}^{9} 4 u d u = \frac{152}{3}

= \frac{1}{25} (\frac{422}{5} - \frac{152}{3})

Simplify

\frac{1}{25} (\frac{422}{5} - \frac{152}{3}) : \frac{506}{375}

= \frac{506}{375}

\frac{d}{d x} (cos (x) ln (x^{2} + 1) - ln (x^{4} + 1))

y^{^{'}} = - y^{3}, y (0) = a

1 + x^{2}

\frac{d}{d x} (\frac{e ^{x}}{9 x ^{2}})

x \to - 1 lim (\frac{x + 1}{x ^{2} + x})

What is the integral from 0 to 1 of xsqrt(4+5x) ?
The integral from 0 to 1 of xsqrt(4+5x) is 506/375