What is the integral of sqrt(6-y) ?

The integral of sqrt(6-y) is -2/3 (6-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(6-y)

\int 6 - y d y

- \frac{2}{3} (6 - y)^{\frac{3}{2}} + C

Solution steps

\int 6 - 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 = 6 - y

= - \frac{2}{3} (6 - y)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (6 - y)^{\frac{3}{2}} + C

\int tan^{3} (7 x + 8) d x

\int 4 - (x)^{2} d x

\int (2 + x + 2 x^{2} + e^{x}) d x

\int x \cdot cos (x) \cdot sin (x) d x

\int \frac{( 4 + x ^{2} ) ^{\frac{1}{2}}}{x ^{6}} d x

What is the integral of sqrt(6-y) ?
The integral of sqrt(6-y) is -2/3 (6-y)^{3/2}+C