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

The integral of (1-(x^2)/4-(y^2)/9) is x-(y^2)/9 x-(x^3)/(12)+C

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

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

x - \frac{y ^{2}}{9} x - \frac{x ^{3}}{12} + C

Solution steps

\int 1 - \frac{x ^{2}}{4} - \frac{y ^{2}}{9} 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

= \int 1 d x - \int \frac{y ^{2}}{9} d x - \int \frac{x ^{2}}{4} d x

\int 1 d x = x

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

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

= x - \frac{y ^{2}}{9} x - \frac{x ^{3}}{12}

Add a constant to the solution

= x - \frac{y ^{2}}{9} x - \frac{x ^{3}}{12} + C

x \to 0 lim (\frac{( \frac{1}{2} ) ^{x} - 1}{x})

f (x) = sec (2 x)

\int e^{x + e^{x}} d x

in v erse l a pl a ce \frac{1}{s ^{2}} - \frac{48}{s ^{5}}

\int \frac{1}{x x ^{2} - 2} d x

What is the integral of (1-(x^2)/4-(y^2)/9) ?
The integral of (1-(x^2)/4-(y^2)/9) is x-(y^2)/9 x-(x^3)/(12)+C