What is the integral from 1 to 2 of (4x^3-9x^2y^2) ?

The integral from 1 to 2 of (4x^3-9x^2y^2) is 4x^3-21x^2

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integral from 1 to 2 of (4x^3-9x^2y^2)

\int_{1}^{2} (4 x^{3} - 9 x^{2} y^{2}) d y

4 x^{3} - 21 x^{2}

Solution steps

\int_{1}^{2} 4 x^{3} - 9 x^{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_{1}^{2} 4 x^{3} d y - \int_{1}^{2} 9 x^{2} y^{2} d y

\int_{1}^{2} 4 x^{3} d y = 4 x^{3}

\int_{1}^{2} 9 x^{2} y^{2} d y = 21 x^{2}

= 4 x^{3} - 21 x^{2}

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What is the integral from 1 to 2 of (4x^3-9x^2y^2) ?
The integral from 1 to 2 of (4x^3-9x^2y^2) is 4x^3-21x^2