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

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

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integral of 3/2 (1-x)^2

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1 - x}{x} d x

\int - 9 x^{2} d x

\int x - \frac{1}{3} d x

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

\int e^{4 x} cos (- 1 x) d x

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