What is the integral of sqrt(6-x) ?

The integral of sqrt(6-x) is -2/3 (6-x)^{3/2}+C

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integral of sqrt(6-x)

\int 6 - x d x

- \frac{2}{3} (6 - x)^{\frac{3}{2}} + C

Solution steps

\int 6 - 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 = 6 - x

= - \frac{2}{3} (6 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (6 - x)^{\frac{3}{2}} + C

\int \frac{( arctan ( x ) )}{( 1 + x ^{2} )} d x

f (x) = 3 x^{2} - 5

x \to 4 - lim (f (x))

\int 1 + 4 x^{2} x d x

\frac{\partial}{\partial x} (9 (x + y)^{2})

What is the integral of sqrt(6-x) ?
The integral of sqrt(6-x) is -2/3 (6-x)^{3/2}+C