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

The integral of (4y-y^2)^2 is (16y^3)/3-2y^4+(y^5)/5+C

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

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

\frac{16 y ^{3}}{3} - 2 y^{4} + \frac{y ^{5}}{5} + C

Solution steps

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

Expand

(4 y - y^{2})^{2} : 16 y^{2} - 8 y^{3} + y^{4}

= \int 16 y^{2} - 8 y^{3} + y^{4} 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 16 y^{2} d y - \int 8 y^{3} d y + \int y^{4} d y

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

\int 8 y^{3} d y = 2 y^{4}

\int y^{4} d y = \frac{y ^{5}}{5}

= \frac{16 y ^{3}}{3} - 2 y^{4} + \frac{y ^{5}}{5}

Add a constant to the solution

= \frac{16 y ^{3}}{3} - 2 y^{4} + \frac{y ^{5}}{5} + C

e^{x + y}

\int sin (\frac{x}{5}) sin (\frac{x}{2}) d x

\int sin (x) \cdot cos (x) d x

y = x^{3} + 4 x, (1, 5)

x \to - 10 lim (∣ x + 10 ∣ + 3)

What is the integral of (4y-y^2)^2 ?
The integral of (4y-y^2)^2 is (16y^3)/3-2y^4+(y^5)/5+C