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 \(\frac{4}{3\sqrt3}\text{ units}^2\).

Since \(y=\sqrt{x}\), rewrite as \(x=y^2\).

The area between the curve and the y-axis is

\(\displaystyle \int_1^4 y^2\,dy\).

\(\displaystyle \int y^2\,dy=\frac{y^3}{3}\).

So

\(\left[\frac{y^3}{3}\right]_1^4=\frac{64-1}{3}=21\).

Therefore, the area is \(21\text{ units}^2\).

The curve crosses the x-axis at \(x=0\).

So the total area is

\(\displaystyle \int_{-2}^{0}(-x^3)\,dx+\int_0^1 x^3\,dx\).

Now

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

So

\(\displaystyle \int_{-2}^{0}(-x^3)\,dx=4\)

and

\(\displaystyle \int_0^1 x^3\,dx=\frac14\).

Total area \(=4+\frac14=\frac{17}{4}\).

The curve meets the y-axis when \(x=0\):

\(y^2-4=0\Rightarrow y=\pm2\).

The area between the y-axis and the curve is

\(\displaystyle \int_{-2}^{2}(0-(y^2-4))\,dy=\int_{-2}^{2}(4-y^2)\,dy\).

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

So

\(\left[4y-\frac{y^3}{3}\right]_{-2}^{2}=\frac{32}{3}\).

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

The area is

\(\displaystyle \int_4^{16}3\sqrt{x}\,dx=3\int_4^{16}x^{1/2}\,dx\).

\(\displaystyle \int x^{1/2}\,dx=\frac{2}{3}x^{3/2}\).

So

\(3\left[\frac{2}{3}x^{3/2}\right]_4^{16}=2(16^{3/2}-4^{3/2})\).

\(16^{3/2}=64\) and \(4^{3/2}=8\).

So the area is \(2(64-8)=112\).

When \(y=0\), \(x=6\).

When \(x=0\), \(y=6\).

So the region is a triangle with base 6 and height 6.

Area \(=\frac12\times6\times6=18\).

Checking by integration:

\(y=6-x\)

\(\displaystyle \int_0^6(6-x)\,dx=18\).

Find the roots:

\(x^2-2x-3=0\Rightarrow (x-3)(x+1)=0\).

So the curve meets the x-axis at \(x=-1\) and \(x=3\).

Between these values the curve is below the x-axis, so

\(\displaystyle \text{Area}=\int_{-1}^{3}(-x^2+2x+3)\,dx\).

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

So

\(\left[-\frac{x^3}{3}+x^2+3x\right]_{-1}^{3}=\frac{32}{3}\).

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

Expand first:

\(y=x^3-4x^2+3x\).

The roots are \(x=0,1,3\).

So the total area is the area on \([0,1]\) plus the positive version of the area on \([1,3]\).

Antiderivative:

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

From 0 to 1, the area is \(\frac{5}{12}\).

From 1 to 3, the integral is \(-\frac{8}{3}\), so the area there is \(\frac{8}{3}\).

Total area \(=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}\).

The curve crosses the x-axis at \(x=2\).

So split the area into two parts:

\(\displaystyle \int_1^2(4-x^2)\,dx+\int_2^3(x^2-4)\,dx\).

For the first part:

\(\displaystyle \left[4x-\frac{x^3}{3}\right]_1^2=\frac{5}{3}\).

For the second part:

\(\displaystyle \left[\frac{x^3}{3}-4x\right]_2^3=\frac{7}{3}\).

Total area \(=\frac{5}{3}+\frac{7}{3}=4\).

The area is

\(\displaystyle \int_0^4(2x+1)\,dx\).

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

So

\(\left[x^2+x\right]_0^4=16+4=20\).

Therefore, the area is \(20\text{ units}^2\).

First find where the curve meets the x-axis:

\(5-2\sqrt{x}=0\Rightarrow \sqrt{x}=\frac52\Rightarrow x=\frac{25}{4}\).

So the area is

\(\displaystyle \int_0^{25/4}(5-2\sqrt{x})\,dx\).

\(\displaystyle \int(5-2x^{1/2})\,dx=5x-\frac{4}{3}x^{3/2}\).

Now \(\left(\frac{25}{4}\right)^{3/2}=\left(\frac52\right)^3=\frac{125}{8}\).

So

\(\frac{125}{4}-\frac{4}{3}\cdot\frac{125}{8}=\frac{125}{4}-\frac{125}{6}=\frac{125}{12}\).

Therefore, the area is \(\frac{125}{12}\text{ units}^2\).

The curve meets the y-axis where \(x=0\):

\(6-y^2=0\Rightarrow y=\pm\sqrt6\).

Using symmetry,

\(\displaystyle \text{Area}=2\int_0^{\sqrt6}(6-y^2)\,dy\).

\(\displaystyle \int(6-y^2)\,dy=6y-\frac{y^3}{3}\).

So

\(2\left[6y-\frac{y^3}{3}\right]_0^{\sqrt6}=8\sqrt6\).

Therefore, the area is \(8\sqrt6\text{ units}^2\).

The area is

\(\displaystyle \int_1^4\frac{1}{x}\,dx\).

\(\displaystyle \int\frac{1}{x}\,dx=\ln x\).

So

\(\left[\ln x\right]_1^4=\ln4-\ln1=\ln4\).

Therefore, the area is \(\ln4\text{ units}^2\).

Find where the curve meets the y-axis:

\(2y-y^2=0\Rightarrow y(2-y)=0\).

So \(y=0\) and \(y=2\).

The area is

\(\displaystyle \int_0^2(2y-y^2)\,dy\).

\(\displaystyle \int(2y-y^2)\,dy=y^2-\frac{y^3}{3}\).

So

\(\left[y^2-\frac{y^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}\).

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

Find the points of intersection:

\(x^2=4x\Rightarrow x(x-4)=0\), so \(x=0\) and \(x=4\).

On \([0,4]\), the line \(y=4x\) lies above the curve \(y=x^2\).

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 points of intersection:

\(x^2=4x+5\Rightarrow x^2-4x-5=0\Rightarrow (x-5)(x+1)=0\).

So \(x=-1\) and \(x=5\).

Between these values, the line \(y=4x+5\) is above the parabola \(y=x^2\).

So the area is

\(\displaystyle \int_{-1}^{5}(4x+5-x^2)\,dx\).

\(\displaystyle \int(4x+5-x^2)\,dx=2x^2+5x-\frac{x^3}{3}\).

So

\(\left[2x^2+5x-\frac{x^3}{3}\right]_{-1}^{5}=36\).

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

Find the points of intersection:

\(x=x^3\Rightarrow x(x^2-1)=0\), so \(x=-1,0,1\).

The curves form two equal regions by symmetry.

On \([0,1]\), \(y=x\) lies above \(y=x^3\).

So the total area 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 \(\frac12\text{ units}^2\).

Find the points of intersection:

\(2x^2=8x\Rightarrow 2x(x-4)=0\), so \(x=0\) and \(x=4\).

On \([0,4]\), \(y=8x\) lies above \(y=2x^2\).

So the area is

\(\displaystyle \int_0^4 (8x-2x^2)\,dx\).

\(\displaystyle \int(8x-2x^2)\,dx=4x^2-\frac{2x^3}{3}\).

So

\(\left[4x^2-\frac{2x^3}{3}\right]_0^4=64-\frac{128}{3}=\frac{64}{3}\).

Therefore, the area is \(\frac{64}{3}\text{ units}^2\).

Find the points of intersection:

\(9-x^2=x+3\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0\).

So \(x=-3\) and \(x=2\).

On \([-3,2]\), \(y=9-x^2\) lies above \(y=x+3\).

So the area is

\(\displaystyle \int_{-3}^{2}(9-x^2-(x+3))\,dx=\int_{-3}^{2}(6-x-x^2)\,dx\).

\(\displaystyle \int(6-x-x^2)\,dx=6x-\frac{x^2}{2}-\frac{x^3}{3}\).

So

\(\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-3}^{2}=\frac{125}{6}\).

Therefore, the area is \(\frac{125}{6}\text{ units}^2\).

Find the points of intersection:

\(x^2+1=5-x^2\Rightarrow 2x^2=4\Rightarrow x=\pm\sqrt2\).

On this interval, \(y=5-x^2\) lies above \(y=x^2+1\).

So the area is

\(\displaystyle \int_{-\sqrt2}^{\sqrt2}\big((5-x^2)-(x^2+1)\big)\,dx=\int_{-\sqrt2}^{\sqrt2}(4-2x^2)\,dx\).

Using symmetry,

\(\displaystyle 2\int_0^{\sqrt2}(4-2x^2)\,dx\).

This gives \(\frac{16\sqrt2}{3}\).

Therefore, the area is \(\frac{16\sqrt2}{3}\text{ units}^2\).

Find the points of intersection:

\(\sqrt{x}=x^2\).

For \(x\ge0\), this gives \(x^{1/2}=x^2\Rightarrow x=0\) or \(x=1\).

On \([0,1]\), \(\sqrt{x}\) lies above \(x^2\).

So the area is

\(\displaystyle \int_0^1 (x^{1/2}-x^2)\,dx\).

\(\displaystyle \int x^{1/2}\,dx=\frac{2}{3}x^{3/2}\) and \(\displaystyle \int x^2\,dx=\frac{x^3}{3}\).

So the area is

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

Find the points of intersection:

\(2\sqrt{x}=x\).

For \(x\ge0\), this gives \(x=0\) or \(x=4\).

On \([0,4]\), \(2\sqrt{x}\) lies above \(x\).

So the area is

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

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

So

\(\left[\frac{4}{3}x^{3/2}-\frac{x^2}{2}\right]_0^4=\frac{32}{3}-8=\frac{8}{3}\).

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

Find the points of intersection:

\(y^2=2y\Rightarrow y(y-2)=0\), so \(y=0\) and \(y=2\).

For \(0\le y\le2\), \(x=2y\) lies to the right of \(x=y^2\).

So the area is

\(\displaystyle \int_0^2 (2y-y^2)\,dy\).

\(\displaystyle \int (2y-y^2)\,dy=y^2-\frac{y^3}{3}\).

So

\(\left[y^2-\frac{y^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}\).

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

Find the points of intersection:

\(y+2=y^2\Rightarrow y^2-y-2=0\Rightarrow (y-2)(y+1)=0\).

So \(y=-1\) and \(y=2\).

For \(-1\le y\le2\), the line \(x=y+2\) lies to the right of the parabola \(x=y^2\).

So the area is

\(\displaystyle \int_{-1}^{2}(y+2-y^2)\,dy\).

\(\displaystyle \int(y+2-y^2)\,dy=\frac{y^2}{2}+2y-\frac{y^3}{3}\).

So

\(\left[\frac{y^2}{2}+2y-\frac{y^3}{3}\right]_{-1}^{2}=\frac{9}{2}\).

Therefore, the area is \(\frac{9}{2}\text{ units}^2\).

Find the points of intersection:

\(3x^2=12-x^2\Rightarrow 4x^2=12\Rightarrow x=\pm\sqrt3\).

On this interval, \(y=12-x^2\) lies above \(y=3x^2\).

So the area is

\(\displaystyle \int_{-\sqrt3}^{\sqrt3}(12-4x^2)\,dx\).

Using symmetry,

\(\displaystyle 2\int_0^{\sqrt3}(12-4x^2)\,dx\).

This gives \(16\sqrt3\).

Therefore, the area is \(16\sqrt3\text{ units}^2\).

Find the points of intersection:

\(x^3=4x\Rightarrow x(x^2-4)=0\).

So \(x=-2,0,2\).

By symmetry, the total area is twice the area on \([0,2]\).

On \([0,2]\), \(y=4x\) lies above \(y=x^3\).

So the total area is

\(\displaystyle 2\int_0^2(4x-x^3)\,dx\).

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

So

\(2\left[2x^2-\frac{x^4}{4}\right]_0^2=2(8-4)=8\).

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

Find the points of intersection:

\(x+1=x^2+1\Rightarrow x=x^2\Rightarrow x(x-1)=0\).

So \(x=0\) and \(x=1\).

On \([0,1]\), \(y=x+1\) lies above \(y=x^2+1\).

So the area is

\(\displaystyle \int_0^1[(x+1)-(x^2+1)]\,dx=\int_0^1(x-x^2)\,dx\).

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

So

\(\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac16\).

Therefore, the area is \(\frac16\text{ units}^2\).

Find the points of intersection:

\(7x-x^2=x\Rightarrow 6x-x^2=0\Rightarrow x(6-x)=0\).

So \(x=0\) and \(x=6\).

On \([0,6]\), \(y=7x-x^2\) lies above \(y=x\).

So the area is

\(\displaystyle \int_0^6[(7x-x^2)-x]dx=\int_0^6(6x-x^2)dx\).

\(\displaystyle \int(6x-x^2)dx=3x^2-\frac{x^3}{3}\).

So

\(\left[3x^2-\frac{x^3}{3}\right]_0^6=108-72=36\).

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

Find the points of intersection:

\(x^{1/3}=x^2\).

For real values, this gives \(x=0\) and \(x=1\).

On \([0,1]\), \(x^{1/3}\) lies above \(x^2\).

So the area is

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

\(\displaystyle \int x^{1/3}\,dx=\frac{3}{4}x^{4/3}\).

So

\(\left[\frac{3}{4}x^{4/3}-\frac{x^3}{3}\right]_0^1=\frac34-\frac13=\frac{5}{12}\).

Therefore, the area is \(\frac{5}{12}\text{ units}^2\).

Find the points of intersection:

\(x^{1/2}=x^{3/2}\Rightarrow x^{1/2}(1-x)=0\).

So \(x=0\) and \(x=1\).

On \([0,1]\), \(x^{1/2}\) lies above \(x^{3/2}\).

So the area is

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

\(\displaystyle \int x^{1/2}\,dx=\frac{2}{3}x^{3/2}\), and \(\displaystyle \int x^{3/2}\,dx=\frac{2}{5}x^{5/2}\).

So the area is

\(\left[\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_0^1=\frac{2}{3}-\frac{2}{5}=\frac{4}{15}\).

Find the points of intersection:

\(x^2=x^4\Rightarrow x^2(1-x^2)=0\).

So \(x=-1,0,1\).

By symmetry, the total area is twice the area on \([0,1]\).

On \([0,1]\), \(x^2\) lies above \(x^4\).

So

\(\displaystyle \text{Area}=2\int_0^1(x^2-x^4)\,dx\).

\(\displaystyle \int(x^2-x^4)dx=\frac{x^3}{3}-\frac{x^5}{5}\).

So

\(2\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=2\left(\frac13-\frac15\right)=\frac{4}{15}\).

Therefore, the total area is \(\frac{4}{15}\text{ units}^2\).

Find the points of intersection:

\(9x-x^2=2x^2\Rightarrow 9x-3x^2=0\Rightarrow 3x(3-x)=0\).

So \(x=0\) and \(x=3\).

On \([0,3]\), \(9x-x^2\) lies above \(2x^2\).

So the area is

\(\displaystyle \int_0^3(9x-3x^2)\,dx\).

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

So

\(\left[\frac{9x^2}{2}-x^3\right]_0^3=\frac{81}{2}-27=\frac{27}{2}\).

Therefore, the area is \(\frac{27}{2}\text{ units}^2\).

Find the points of intersection:

\(x=x^4\Rightarrow x(x^3-1)=0\), so \(x=0\) and \(x=1\).

On \([0,1]\), \(x\) lies above \(x^4\).

So the area is

\(\displaystyle \int_0^1(x-x^4)\,dx\).

\(\displaystyle \int(x-x^4)dx=\frac{x^2}{2}-\frac{x^5}{5}\).

So

\(\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=\frac12-\frac15=\frac{3}{10}\).

Therefore, the area is \(\frac{3}{10}\text{ units}^2\).

Find the points of intersection:

\(x^3+2x=5x\Rightarrow x^3-3x=0\Rightarrow x(x^2-3)=0\).

So \(x=-\sqrt3,0,\sqrt3\).

By symmetry, the total area is twice the area on \([0,\sqrt3]\).

On \([0,\sqrt3]\), \(5x\) lies above \(x^3+2x\), so the difference is \(3x-x^3\).

Thus

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

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

So

\(2\left[\frac{3x^2}{2}-\frac{x^4}{4}\right]_0^{\sqrt3}=2\left(\frac{9}{2}-\frac{9}{4}\right)=\frac{9}{2}\).

Therefore, the total area is \(\frac{9}{2}\text{ units}^2\).