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

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

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integral of sqrt(4-t)

\int 4 - t d t

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

Solution steps

\int 4 - t d t

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 - t

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

Add a constant to the solution

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

\int_{0}^{2} (x^{2} - 2)^{4} x d x

y = 4 x, y = \frac{5}{4} x, y = 21 - x^{2}

x \to - 1 - lim (\frac{x ^{3}}{3 x + 3})

3 x^{4}

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

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