What is the integral of sqrt(9-x) ?

The integral of sqrt(9-x) is -2/3 (9-x)^{3/2}+C

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integral of sqrt(9-x)

\int 9 - x d x

- \frac{2}{3} (9 - x)^{\frac{3}{2}} + C

Solution steps

\int 9 - x d x

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 - x

= - \frac{2}{3} (9 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (9 - x)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (3 x^{3})

f (x) = - \frac{4}{x ^{2}}

y = 8 e^{x} cos (x)

\int (sin (2 x))^{2} cos (2 x) d x

- 4 sin (4 t)

What is the integral of sqrt(9-x) ?
The integral of sqrt(9-x) is -2/3 (9-x)^{3/2}+C