What is the integral from 0 to 4 of x^2sqrt(2x+1) ?

The integral from 0 to 4 of x^2sqrt(2x+1) is 5884/105

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integral from 0 to 4 of x^2sqrt(2x+1)

\int_{0}^{4} x^{2} 2 x + 1 d x

\frac{5884}{105}

Decimal

56.03809 \dots

Solution steps

\int_{0}^{4} x^{2} 2 x + 1 d x

Apply u-substitution

: \int_{1}^{9} \frac{u ( u - 1 ) ^{2}}{8} d u

= \int_{1}^{9} \frac{u ( u - 1 ) ^{2}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{9} u (u - 1)^{2} d u

Expand

u (u - 1)^{2} : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

= \frac{1}{8} \cdot \int_{1}^{9} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + 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}{8} (\int_{1}^{9} u^{\frac{5}{2}} d u - \int_{1}^{9} 2 u^{\frac{3}{2}} d u + \int_{1}^{9} u d u)

\int_{1}^{9} u^{\frac{5}{2}} d u = \frac{4372}{7}

\int_{1}^{9} 2 u^{\frac{3}{2}} d u = \frac{968}{5}

\int_{1}^{9} u d u = \frac{52}{3}

= \frac{1}{8} (\frac{4372}{7} - \frac{968}{5} + \frac{52}{3})

Simplify

\frac{1}{8} (\frac{4372}{7} - \frac{968}{5} + \frac{52}{3}) : \frac{5884}{105}

= \frac{5884}{105}

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What is the integral from 0 to 4 of x^2sqrt(2x+1) ?
The integral from 0 to 4 of x^2sqrt(2x+1) is 5884/105