What is the integral of 9x(1-x) ?

The integral of 9x(1-x) is 9((x^2)/2-(x^3)/3)+C

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integral of 9x(1-x)

\int 9 x (1 - x) d x

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

Solution steps

\int 9 x (1 - x) d x

Take the constant out:

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

= 9 \cdot \int x (1 - x) d x

Expand

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

= 9 \cdot \int 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

= 9 (\int x d x - \int x^{2} d x)

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

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

= 9 (\frac{x ^{2}}{2} - \frac{x ^{3}}{3})

Add a constant to the solution

= 9 (\frac{x ^{2}}{2} - \frac{x ^{3}}{3}) + C

\int_{0}^{π} - cos (x) d x

cos (x) \frac{d y}{d x} + y sin (x) = 1

x \to 0 lim (lo g_{\frac{1}{2}} (x))

\frac{x ^{3}}{x + 1}

\int 18 e^{x} d x

What is the integral of 9x(1-x) ?
The integral of 9x(1-x) is 9((x^2)/2-(x^3)/3)+C