What is the integral from 1 to 2y of 2xy ?

The integral from 1 to 2y of 2xy is 4y^3-y

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integral from 1 to 2y of 2xy

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

4 y^{3} - y

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{2 y}

= 2 y [\frac{x ^{2}}{2}]_{1}^{2 y}

Compute the boundaries:

2 y^{2} - \frac{1}{2}

= 2 y (2 y^{2} - \frac{1}{2})

Simplify

= 4 y^{3} - y

\int_{0}^{1} \frac{x + 2}{x ^{2} - x - 2} d x

\int_{- 1}^{3} ∣ x - 1 ∣ d x

\int_{- 2}^{0} cos (\frac{nπ x}{2}) d x

\int_{- 4}^{1} - x^{2} - 3 x d x

\int_{2}^{4} - x^{2} + 2 x + 8 d x

What is the integral from 1 to 2y of 2xy ?
The integral from 1 to 2y of 2xy is 4y^3-y