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

The integral of sqrt(16-x) is -2/3 (16-x)^{3/2}+C

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integral of sqrt(16-x)

\int 16 - x d x

- \frac{2}{3} (16 - x)^{\frac{3}{2}} + C

Solution steps

\int 16 - 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 = 16 - x

= - \frac{2}{3} (16 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (16 - x)^{\frac{3}{2}} + C

\int \frac{x ^{5}}{x ^{6} - 4} d x

\frac{d}{d x} (csc (x) tan (x))

4 sin (t) \frac{d y}{d t} + 4 y cos (t) = cos (t) sin^{4} (t)

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} + x^{2}

x \to 0 lim (x^{2} - 2)

What is the integral of sqrt(16-x) ?
The integral of sqrt(16-x) is -2/3 (16-x)^{3/2}+C