What is the integral of 3(1-x)^2 ?

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

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integral of 3(1-x)^2

\int 3 (1 - x)^{2} d x

3 x - 3 x^{2} + x^{3} + C

Solution steps

\int 3 (1 - x)^{2} d x

Take the constant out:

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

= 3 \cdot \int (1 - x)^{2} d x

Expand

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

= 3 \cdot \int 1 - 2 x + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d x = x

\int 2 x d x = x^{2}

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

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

Simplify

3 (x - x^{2} + \frac{x ^{3}}{3}) : 3 x - 3 x^{2} + x^{3}

= 3 x - 3 x^{2} + x^{3}

Add a constant to the solution

= 3 x - 3 x^{2} + x^{3} + C

\int x^{\frac{1}{2}} + 6 d x

\int_{0}^{2} x^{3} (2 - x)^{2} d x

x \to 0 lim (\frac{e ^{x}}{ln ( x )})

\frac{3 x ^{2} + x}{- x ^{2} + 3 x - 3}

17 - x^{2}