What is the integral from 0 to 21 of x(x+4)^{1/2} ?

The integral from 0 to 21 of x(x+4)^{1/2} is 4626/5

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integral from 0 to 21 of x(x+4)^{1/2}

\int_{0}^{21} x (x + 4)^{\frac{1}{2}} d x

\frac{4626}{5}

Decimal Notation

925.2

Solution steps

\int_{0}^{21} x (x + 4)^{\frac{1}{2}} d x

Apply u-substitution

: \int_{2}^{5} 2 (u^{2} - 4) u^{2} d u

= \int_{2}^{5} 2 (u^{2} - 4) u^{2} d u

Take the constant out:

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

= 2 \cdot \int_{2}^{5} (u^{2} - 4) u^{2} d u

Expand

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

= 2 \cdot \int_{2}^{5} u^{4} - 4 u^{2} 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

= 2 (\int_{2}^{5} u^{4} d u - \int_{2}^{5} 4 u^{2} d u)

\int_{2}^{5} u^{4} d u = \frac{3093}{5}

\int_{2}^{5} 4 u^{2} d u = 156

= 2 (\frac{3093}{5} - 156)

Simplify

2 (\frac{3093}{5} - 156) : \frac{4626}{5}

= \frac{4626}{5}

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What is the integral from 0 to 21 of x(x+4)^{1/2} ?
The integral from 0 to 21 of x(x+4)^{1/2} is 4626/5