What is the integral of (y^2+4y-8) ?

The integral of (y^2+4y-8) is (y^3)/3+2y^2-8y+C

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integral of (y^2+4y-8)

\int (y^{2} + 4 y - 8) d y

\frac{y ^{3}}{3} + 2 y^{2} - 8 y + C

Solution steps

\int y^{2} + 4 y - 8 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 y^{2} d y + \int 4 y d y - \int 8 d y

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

\int 4 y d y = 2 y^{2}

\int 8 d y = 8 y

= \frac{y ^{3}}{3} + 2 y^{2} - 8 y

Add a constant to the solution

= \frac{y ^{3}}{3} + 2 y^{2} - 8 y + C

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What is the integral of (y^2+4y-8) ?
The integral of (y^2+4y-8) is (y^3)/3+2y^2-8y+C