What is the integral of-(4y-x^2)^2 ?

The integral of-(4y-x^2)^2 is -x^4y+4x^2y^2-(16y^3)/3+C

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integral of-(4y-x^2)^2

\int - (4 y - x^{2})^{2} d y

- x^{4} y + 4 x^{2} y^{2} - \frac{16 y ^{3}}{3} + C

Solution steps

\int - (4 y - x^{2})^{2} d y

Take the constant out:

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

= - \int (- x^{2} + 4 y)^{2} d y

Expand

(- x^{2} + 4 y)^{2} : x^{4} - 8 x^{2} y + 16 y^{2}

= - \int x^{4} - 8 x^{2} y + 16 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 x^{4} d y - \int 8 x^{2} y d y + \int 16 y^{2} d y)

\int x^{4} d y = x^{4} y

\int 8 x^{2} y d y = 4 x^{2} y^{2}

\int 16 y^{2} d y = \frac{16 y ^{3}}{3}

= - (x^{4} y - 4 x^{2} y^{2} + \frac{16 y ^{3}}{3})

Simplify

= - x^{4} y + 4 x^{2} y^{2} - \frac{16 y ^{3}}{3}

Add a constant to the solution

= - x^{4} y + 4 x^{2} y^{2} - \frac{16 y ^{3}}{3} + C

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What is the integral of-(4y-x^2)^2 ?
The integral of-(4y-x^2)^2 is -x^4y+4x^2y^2-(16y^3)/3+C