What is the integral of (-y^2+1)-(2y^2-2) ?

The integral of (-y^2+1)-(2y^2-2) is -y^3+3y+C

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integral of (-y^2+1)-(2y^2-2)

\int (- y^{2} + 1) - (2 y^{2} - 2) d y

- y^{3} + 3 y + C

Solution steps

\int - y^{2} + 1 - (2 y^{2} - 2) d y

Simplify

= \int - 3 y^{2} + 3 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 3 y^{2} d y + \int 3 d y

\int 3 y^{2} d y = y^{3}

\int 3 d y = 3 y

= - y^{3} + 3 y

Add a constant to the solution

= - y^{3} + 3 y + C

sec^{6} (x)

t \frac{d y}{d t} = t^{3} + 12 t^{3} y

\int x^{2} - x d x

x \to + 0 lim (9 x - 4)

x \to - 1 - lim (2)

What is the integral of (-y^2+1)-(2y^2-2) ?
The integral of (-y^2+1)-(2y^2-2) is -y^3+3y+C