What is the integral of 6x(9-x) ?

The integral of 6x(9-x) is 6((9x^2)/2-(x^3)/3)+C

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integral of 6x(9-x)

\int 6 x (9 - x) d x

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

Solution steps

\int 6 x (9 - x) d x

Take the constant out:

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

= 6 \cdot \int x (9 - x) d x

Expand

x (9 - x) : 9 x - x^{2}

= 6 \cdot \int 9 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

= 6 (\int 9 x d x - \int x^{2} d x)

\int 9 x d x = \frac{9 x ^{2}}{2}

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

= 6 (\frac{9 x ^{2}}{2} - \frac{x ^{3}}{3})

Add a constant to the solution

= 6 (\frac{9 x ^{2}}{2} - \frac{x ^{3}}{3}) + C

\frac{d}{d x} (lo g_{10} (x^{3} + 9))

2, [0, \frac{π}{3}]

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

x \to 0 lim (x^{s i n (3 x)})

Y = (3 - X^{2}) (X^{3} - X + 1), (2, 1)

What is the integral of 6x(9-x) ?
The integral of 6x(9-x) is 6((9x^2)/2-(x^3)/3)+C