What is the integral of 1/9 (4-x^2) ?

The integral of 1/9 (4-x^2) is 1/9 (4x-(x^3)/3)+C

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integral of 1/9 (4-x^2)

\int \frac{1}{9} (4 - x^{2}) d x

\frac{1}{9} (4 x - \frac{x ^{3}}{3}) + C

Solution steps

\int \frac{1}{9} (4 - x^{2}) d x

Take the constant out:

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

= \frac{1}{9} \cdot \int 4 - 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{1}{9} (\int 4 d x - \int x^{2} d x)

\int 4 d x = 4 x

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

= \frac{1}{9} (4 x - \frac{x ^{3}}{3})

Add a constant to the solution

= \frac{1}{9} (4 x - \frac{x ^{3}}{3}) + C

y = x^{3} - x^{2} - 6 x, x

\frac{\partial}{\partial x} (y^{5} sin (5 x))

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

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\int 8 x^{5} e^{5 x^{6}} d x

What is the integral of 1/9 (4-x^2) ?
The integral of 1/9 (4-x^2) is 1/9 (4x-(x^3)/3)+C