What is the integral from 1 to y of (2xy+y^2) ?

The integral from 1 to y of (2xy+y^2) is 2y^3-y^2-y

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integral from 1 to y of (2xy+y^2)

\int_{1}^{y} (2 x y + y^{2}) d x

2 y^{3} - y^{2} - y

Solution steps

\int_{1}^{y} 2 x y + y^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{y} y^{2} d x + \int_{1}^{y} 2 y x d x

\int_{1}^{y} y^{2} d x = y^{3} - y^{2}

\int_{1}^{y} 2 y x d x = y (y^{2} - 1)

= y^{3} - y^{2} + y (y^{2} - 1)

Simplify

y^{3} - y^{2} + y (y^{2} - 1) : 2 y^{3} - y^{2} - y

= 2 y^{3} - y^{2} - y

\int_{- 1}^{2} x^{3} + 8 d x

\int_{- π}^{π} ∣ t ∣ d t

\int_{0}^{y} x^{2} + y^{2} d x

\int_{0}^{x} 4 + t^{6} d x

\int_{0}^{1} 1 + 8 t^{2} d t

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