What is the integral of (1-t^{-0.52}) ?

The integral of (1-t^{-0.52}) is t-2.08333…t^{0.48}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-t^{-0.52})

\int (1 - t^{- 0.52}) d t

t - 2.08333 \dots t^{0.48} + C

Solution steps

\int 1 - t^{- 0.52} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d t - \int t^{- 0.52} d t

\int 1 d t = t

\int t^{- 0.52} d t = 2.08333 \dots t^{0.48}

= t - 2.08333 \dots t^{0.48}

Add a constant to the solution

= t - 2.08333 \dots t^{0.48} + C

y^{^{'}} = y - 1

\int_{0}^{10} (6 e^{0.2 x} + 2) d x

\int \frac{3 x}{2 x} d x

x \to 4 + lim (\frac{6 x}{x - 4})

What is the integral of (1-t^{-0.52}) ?
The integral of (1-t^{-0.52}) is t-2.08333…t^{0.48}+C