First find where the curve meets the x-axis:

\(x^2-9=0\Rightarrow x=\pm3\).

Between \(x=-3\) and \(x=3\), the curve lies below the x-axis, so the area is

\(\displaystyle \int_{-3}^{3}(9-x^2)\,dx\).

\(\displaystyle \int (9-x^2)\,dx=9x-\frac{x^3}{3}\).

So

\(\left[9x-\frac{x^3}{3}\right]_{-3}^{3}=36\).

Therefore, the area is \(36\text{ units}^2\).

First find the x-intercepts:

\(x(x-4)=0\Rightarrow x=0\text{ and }x=4\).

On \([0,4]\), the curve is below the x-axis, so the area is

\(\displaystyle \int_0^4 (4x-x^2)\,dx\).

\(\displaystyle \int (4x-x^2)\,dx=2x^2-\frac{x^3}{3}\).

So

\(\left[2x^2-\frac{x^3}{3}\right]_0^4=32-\frac{64}{3}=\frac{32}{3}\).

Therefore, the area is \(\frac{32}{3}\text{ units}^2\).

Find the x-intercepts:

\(2-x^2=0\Rightarrow x=\pm\sqrt2\).

The curve is above the x-axis between these values, so

\(\displaystyle \text{Area}=\int_{-\sqrt2}^{\sqrt2}(2-x^2)\,dx\).

Using symmetry,

\(\displaystyle \text{Area}=2\int_0^{\sqrt2}(2-x^2)\,dx\).

\(\displaystyle \int(2-x^2)\,dx=2x-\frac{x^3}{3}\).

So

\(2\left[2x-\frac{x^3}{3}\right]_0^{\sqrt2}=2\left(2\sqrt2-\frac{2\sqrt2}{3}\right)=\frac{8\sqrt2}{3}\).

Therefore, the area is \(\frac{8\sqrt2}{3}\text{ units}^2\).

Factorise:

\(y=x(x-1)(x+1)\), so the roots are \(-1,0,1\).

The curve is above the x-axis on \([-1,0]\) and below on \([0,1]\).

So

\(\displaystyle \text{Area}=\int_{-1}^{0}(x^3-x)\,dx+\int_0^1(x-x^3)\,dx\).

By symmetry, this is

\(\displaystyle 2\int_0^1(x-x^3)\,dx\).

\(\displaystyle \int(x-x^3)\,dx=\frac{x^2}{2}-\frac{x^4}{4}\).

So

\(2\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=2\left(\frac12-\frac14\right)=\frac12\).

Therefore, the total area is \(\frac12\text{ units}^2\).

First find where the curve meets the x-axis:

\(1-3x^2=0\Rightarrow x=\pm\frac{1}{\sqrt3}\).

Using symmetry,

\(\displaystyle \text{Area}=2\int_0^{1/\sqrt3}(1-3x^2)\,dx\).

\(\displaystyle \int(1-3x^2)\,dx=x-x^3\).

So

\(2\left[x-x^3\right]_0^{1/\sqrt3}=2\left(\frac{1}{\sqrt3}-\frac{1}{3\sqrt3}\right)=\frac{4}{3\sqrt3}\).

Therefore, the area is \(\frac{4}{3\sqrt3}\text{ units}^2\).