What is the integral from 0 to 9 of (sqrt(x)-1/3 x) ?

The integral from 0 to 9 of (sqrt(x)-1/3 x) is 9/2

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integral from 0 to 9 of (sqrt(x)-1/3 x)

\int_{0}^{9} (x - \frac{1}{3} x) d x

\frac{9}{2}

Solution steps

\int_{0}^{9} x - \frac{1}{3} x 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

= \int_{0}^{9} x d x - \int_{0}^{9} \frac{1}{3} x d x

\int_{0}^{9} x d x = 18

\int_{0}^{9} \frac{1}{3} x d x = \frac{27}{2}

= 18 - \frac{27}{2}

Simplify

18 - \frac{27}{2} : \frac{9}{2}

= \frac{9}{2}

\int e^{2 x} cos (7 x) d x

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\frac{\partial}{\partial x} (sin (π (2 x - y)))

\int x^{7} d x

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What is the integral from 0 to 9 of (sqrt(x)-1/3 x) ?
The integral from 0 to 9 of (sqrt(x)-1/3 x) is 9/2