What is the integral of 1/(2sqrt(y)) ?

The integral of 1/(2sqrt(y)) is sqrt(y)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2sqrt(y))

\int \frac{1}{2 y} d y

y + C

Solution steps

\int \frac{1}{2 y} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{y} d y

Apply the Power Rule

: 2 y^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 y^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 y^{\frac{1}{2}} : y

= y

Add a constant to the solution

= y + C

e^{2 x}

sec^{9} (x)

y^{^{''}} + 2 y^{^{'}} + y = e^{- x} ln (x)

\frac{d}{d x} (\frac{x ^{6}}{6} - \frac{x ^{5}}{5} + 20)

f (x) = x, g (x) = x, x = 0, x = 2

What is the integral of 1/(2sqrt(y)) ?
The integral of 1/(2sqrt(y)) is sqrt(y)+C