What is the integral of y^2(1-2x) ?

The integral of y^2(1-2x) is (y^3(1-2x))/3+C

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integral of y^2(1-2x)

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

\frac{y ^{3} ( 1 - 2 x )}{3} + C

Solution steps

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

Take the constant out:

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

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

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

= (1 - 2 x) \frac{y ^{3}}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{y ^{3} ( 1 - 2 x )}{3}

Add a constant to the solution

= \frac{y ^{3} ( 1 - 2 x )}{3} + C

\frac{\partial}{\partial x} (y (e^{x} - 1))

x \to 0 lim (x^{2} + 16 x + 25)

k = 0 \sum \infty \frac{2}{k !}

\int (4 x - 3)^{2} d x

x \to 0 + lim ((ln (x))^{x})

What is the integral of y^2(1-2x) ?
The integral of y^2(1-2x) is (y^3(1-2x))/3+C