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

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

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

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

\frac{506}{375}

Decimal Notation

1.34933 \dots

Solution steps

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

\int x^{2} + sin (x) d x

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\frac{d}{d x} (x n (x))

\int \frac{1}{v ^{2} + 3 v - 4} d v

\frac{- 4}{( 7 x ^{5} + 6 ) ^{6}}

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