What is the integral of 3/8 (4x-2x^2) ?

The integral of 3/8 (4x-2x^2) is 1/4 x^2(-x+3)+C

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integral of 3/8 (4x-2x^2)

\int \frac{3}{8} (4 x - 2 x^{2}) d x

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

Solution steps

\int \frac{3}{8} (4 x - 2 x^{2}) d x

Take the constant out:

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

= \frac{3}{8} \cdot \int 4 x - 2 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}{8} (\int 4 x d x - \int 2 x^{2} d x)

\int 4 x d x = 2 x^{2}

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

= \frac{3}{8} (2 x^{2} - \frac{2 x ^{3}}{3})

Simplify

\frac{3}{8} (2 x^{2} - \frac{2 x ^{3}}{3}) : \frac{1}{4} x^{2} (- x + 3)

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

Add a constant to the solution

= \frac{1}{4} x^{2} (- x + 3) + C

What is the integral of 3/8 (4x-2x^2) ?
The integral of 3/8 (4x-2x^2) is 1/4 x^2(-x+3)+C