What is the integral of (-1)/(2sqrt(16-x)) ?

The integral of (-1)/(2sqrt(16-x)) is sqrt(16-x)+C

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integral of (-1)/(2sqrt(16-x))

\int \frac{- 1}{2 16 - x} d x

16 - x + C

Solution steps

\int \frac{- 1}{2 16 - x} d x

Take the constant out:

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

= \frac{- 1}{2} \cdot \int \frac{1}{16 - x} d x

Simplify

= (- \frac{1}{2}) \cdot \int \frac{1}{16 - x} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= (- \frac{1}{2}) \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= (- \frac{1}{2}) (- \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= (- \frac{1}{2}) (- 2 u^{\frac{1}{2}})

Substitute back

u = 16 - x

= (- \frac{1}{2}) (- 2 (16 - x)^{\frac{1}{2}})

Simplify

(- \frac{1}{2}) (- 2 (16 - x)^{\frac{1}{2}}) : 16 - x

= 16 - x

Add a constant to the solution

= 16 - x + C

x \to \infty lim (c e^{x})

\frac{d}{d x} (4 x^{2} + 8)

\int_{0}^{π} cos (x) - sin (x) d x

\frac{\partial}{\partial x} ((x + y)^{\frac{1}{2}})

\frac{\partial}{\partial x} (4 x^{3} + y^{2})

What is the integral of (-1)/(2sqrt(16-x)) ?
The integral of (-1)/(2sqrt(16-x)) is sqrt(16-x)+C