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

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

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integral from 1 to x of (2xy+3y^2)

\int_{1}^{x} (2 x y + 3 y^{2}) d y

2 x^{3} - x - 1

Solution steps

\int_{1}^{x} 2 x y + 3 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_{1}^{x} 3 y^{2} d y + \int_{1}^{x} 2 x y d y

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

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

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

Simplify

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

= 2 x^{3} - x - 1

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

\int_{0}^{2} [x] d x

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