What is the integral of sqrt(4-x) ?

The integral of sqrt(4-x) is -2/3 (4-x)^{3/2}+C

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integral of sqrt(4-x)

\int 4 - x d x

- \frac{2}{3} (4 - x)^{\frac{3}{2}} + C

Solution steps

\int 4 - 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 = 4 - x

= - \frac{2}{3} (4 - x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (4 - x)^{\frac{3}{2}} + C

\frac{d}{d x} (\frac{8}{x ^{2} + 1})

x (x^{2} + 1)^{3}

f (x) = x^{4} - 4 x^{2}, at x = 1

\int 25 - y^{2} d y

3 x - 6, - 2 x + 10, 2, 5

What is the integral of sqrt(4-x) ?
The integral of sqrt(4-x) is -2/3 (4-x)^{3/2}+C