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

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

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integral of ((1-x)^2)/2

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

= \frac{1}{2} \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

= \frac{1}{2} (\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}

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

Add a constant to the solution

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

\int tan^{2} (x) + 1 d x

\int_{3}^{5} \frac{1}{( x - 4 ) ^{2}} d x

\int ln (x) x^{5} d x

\frac{\partial}{\partial x} (\frac{x - 3 y}{2 x y + 3})

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