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

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

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

integral

(1 - x)^{2}

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

Solution steps

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

(1 - x)^{2} = ((1 - x)),

assuming

(1 - x) \geq 0

= \int 1 - x 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

= \int 1 d x - \int x d x

\int 1 d x = x

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

= x - \frac{x ^{2}}{2}

Add a constant to the solution

= x - \frac{x ^{2}}{2} + C

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

x^{2} e^{- x^{2}}

\frac{sin ^{4} ( x )}{cos ^{2} ( x )}

\frac{cos ( x )}{x} + \frac{sin ( x )}{x ^{2}}

\frac{2 x + 4}{x ^{3} - 1}

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