What is the integral of ((9-y^2)^2)/2-y(9-y^2) ?

The integral of ((9-y^2)^2)/2-y(9-y^2) is 81/2 y-3y^3+(y^5}{10}-\frac{9y^2)/2+(y^4)/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((9-y^2)^2)/2-y(9-y^2)

\int \frac{( 9 - y ^{2} ) ^{2}}{2} - y (9 - y^{2}) d y

\frac{81}{2} y - 3 y^{3} + \frac{y ^{5}}{10} - \frac{9 y ^{2}}{2} + \frac{y ^{4}}{4} + C

Solution steps

\int \frac{( 9 - y ^{2} ) ^{2}}{2} - y (9 - y^{2}) d y

Apply the Sum Rule:

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

= \int \frac{( 9 - y ^{2} ) ^{2}}{2} d y - \int y (9 - y^{2}) d y

\int \frac{( 9 - y ^{2} ) ^{2}}{2} d y = \frac{1}{2} (81 y - 6 y^{3} + \frac{y ^{5}}{5})

\int y (9 - y^{2}) d y = \frac{9 y ^{2}}{2} - \frac{y ^{4}}{4}

= \frac{1}{2} (81 y - 6 y^{3} + \frac{y ^{5}}{5}) - (\frac{9 y ^{2}}{2} - \frac{y ^{4}}{4})

Simplify

\frac{1}{2} (81 y - 6 y^{3} + \frac{y ^{5}}{5}) - (\frac{9 y ^{2}}{2} - \frac{y ^{4}}{4}) : \frac{81}{2} y - 3 y^{3} + \frac{y ^{5}}{10} - \frac{9 y ^{2}}{2} + \frac{y ^{4}}{4}

= \frac{81}{2} y - 3 y^{3} + \frac{y ^{5}}{10} - \frac{9 y ^{2}}{2} + \frac{y ^{4}}{4}

Add a constant to the solution

= \frac{81}{2} y - 3 y^{3} + \frac{y ^{5}}{10} - \frac{9 y ^{2}}{2} + \frac{y ^{4}}{4} + C

\int y + 4 d y

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

\int_{1}^{4} - 10 t ln (t) d t

\int 10 x^{5} e^{- x^{3}} d x

\frac{d}{d x} (- 3 x^{2} + 2)

What is the integral of ((9-y^2)^2)/2-y(9-y^2) ?
The integral of ((9-y^2)^2)/2-y(9-y^2) is 81/2 y-3y^3+(y^5}{10}-\frac{9y^2)/2+(y^4)/4+C