What is the integral of (t)^{1/2} ?

The integral of (t)^{1/2} is 2/3 t^{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 (t)^{1/2}

\int (t)^{\frac{1}{2}} d t

\frac{2}{3} t^{\frac{3}{2}} + C

Solution steps

\int t^{\frac{1}{2}} d t

Apply the Power Rule

: \frac{2}{3} t^{\frac{3}{2}}

= \frac{2}{3} t^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} t^{\frac{3}{2}} + C

\int tan (6 x) (sec (6 x))^{2} d x

\int 8 sin (x) d x

\frac{d}{d y} (1 - \frac{3}{y} + x)

\frac{d y}{d x} = - y^{2} sin (x)

\int \frac{1}{x ( x ^{2} - 49 ) ^{\frac{3}{2}}} d x