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

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

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

\int (y + 2) (y - 1) d y

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

Solution steps

\int (y + 2) (y - 1) d y

Expand

(y + 2) (y - 1) : y^{2} + y - 2

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

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

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

\int 2 d y = 2 y

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

Add a constant to the solution

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

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