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

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

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integral of (y^3-y^2+1)

\int (y^{3} - y^{2} + 1) d y

\frac{y ^{4}}{4} - \frac{y ^{3}}{3} + y + C

Solution steps

\int y^{3} - y^{2} + 1 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^{3} d y - \int y^{2} d y + \int 1 d y

\int y^{3} d y = \frac{y ^{4}}{4}

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

\int 1 d y = y

= \frac{y ^{4}}{4} - \frac{y ^{3}}{3} + y

Add a constant to the solution

= \frac{y ^{4}}{4} - \frac{y ^{3}}{3} + y + C

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