What is the integral from 0 to 9 of 6-2sqrt(x) ?

The integral from 0 to 9 of 6-2sqrt(x) is 18

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integral from 0 to 9 of 6-2sqrt(x)

\int_{0}^{9} 6 - 2 x d x

18

Solution steps

\int_{0}^{9} 6 - 2 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} 6 d x - \int_{0}^{9} 2 x d x

\int_{0}^{9} 6 d x = 54

\int_{0}^{9} 2 x d x = 36

= 54 - 36

Simplify

= 18

\int \frac{x}{25 + 16 x ^{4}} d x

\int \frac{x ^{2}}{6 - x} d x

x \to 2 lim (f (x))

y = x^{3} - 9 x^{2} + 18 x, y = - x^{3} + 9 x^{2} - 18 x

\int x sin (6 - x) d x

What is the integral from 0 to 9 of 6-2sqrt(x) ?
The integral from 0 to 9 of 6-2sqrt(x) is 18