What is the integral of (y+4)/2-(y^2)/2 ?

The integral of (y+4)/2-(y^2)/2 is 1/2 ((y^2)/2+4y)-(y^3)/6+C

integral of (y+4)/2-(y^2)/2

\int \frac{y + 4}{2} - \frac{y ^{2}}{2} d y

\frac{1}{2} (\frac{y ^{2}}{2} + 4 y) - \frac{y ^{3}}{6} + C

Solution steps

\int \frac{y + 4}{2} - \frac{y ^{2}}{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{y + 4}{2} d y - \int \frac{y ^{2}}{2} d y

\int \frac{y + 4}{2} d y = \frac{1}{2} (\frac{y ^{2}}{2} + 4 y)

\int \frac{y ^{2}}{2} d y = \frac{y ^{3}}{6}

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

Add a constant to the solution

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

\int \frac{4 ln ( x )}{x} d x

d er i v a t i v e x^{π}

d er i v a t i v e e^{9 t s i n (2 t)}

\int \frac{x - 5}{x ^{2} - x - 2} d x

\frac{\partial}{\partial x} (x^{6} y^{5} - x^{5} y^{4})

What is the integral of (y+4)/2-(y^2)/2 ?
The integral of (y+4)/2-(y^2)/2 is 1/2 ((y^2)/2+4y)-(y^3)/6+C