What is the integral from 0 to y^2 of sqrt(y^3+1) ?

The integral from 0 to y^2 of sqrt(y^3+1) is y^2sqrt(y^3+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to y^2 of sqrt(y^3+1)

\int_{0}^{y^{2}} y^{3} + 1 d x

y^{2} y^{3} + 1

Solution steps

\int_{0}^{y^{2}} y^{3} + 1 d x

Integral of a constant:

\int a d x = a x

= [1 + y^{3} x]_{0}^{y^{2}}

Compute the boundaries:

y^{2} y^{3} + 1

= y^{2} y^{3} + 1

\int_{- 2}^{- 1} 6 x (2 x^{2} - 6)^{3} d x

\int_{0}^{9} 2 t d t

\int_{1}^{3} - x^{2} d x

\int_{1}^{1.5} 1 d x

\int_{0}^{4} (4 - x^{2})^{2} d x

What is the integral from 0 to y^2 of sqrt(y^3+1) ?
The integral from 0 to y^2 of sqrt(y^3+1) is y^2sqrt(y^3+1)