What is the integral from 0 to 9 of 2x(sqrt(x)+x) ?

The integral from 0 to 9 of 2x(sqrt(x)+x) is 3402/5

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integral from 0 to 9 of 2x(sqrt(x)+x)

\int_{0}^{9} 2 x (x + x) d x

\frac{3402}{5}

Decimal Notation

680.4

Solution steps

\int_{0}^{9} 2 x (x + x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{9} x (x + x) d x

Expand

x (x + x) : x^{\frac{3}{2}} + x^{2}

= 2 \cdot \int_{0}^{9} x^{\frac{3}{2}} + x^{2} d x

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_{0}^{9} x^{\frac{3}{2}} d x + \int_{0}^{9} x^{2} d x)

\int_{0}^{9} x^{\frac{3}{2}} d x = \frac{486}{5}

\int_{0}^{9} x^{2} d x = 243

= 2 (\frac{486}{5} + 243)

Simplify

2 (\frac{486}{5} + 243) : \frac{3402}{5}

= \frac{3402}{5}

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What is the integral from 0 to 9 of 2x(sqrt(x)+x) ?
The integral from 0 to 9 of 2x(sqrt(x)+x) is 3402/5