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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(16-y)

\int 16 - y d y

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

Solution steps

\int 16 - y d y

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 - y

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

Add a constant to the solution

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

\int_{0}^{2} - 10 x^{3} x^{2} + 4 d x

6 t \frac{d y}{d t} + y = t^{7}

\frac{2}{x + 2}

f (x) = (- 4 x^{2} + 16)^{2}

\int_{0}^{\frac{π}{4}} ln (sec (x)) d x

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