Use the power rule for integration:

\(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\).

So

\(\int 6x^2\,dx=6\cdot\dfrac{x^3}{3}+C=2x^3+C\).

Rewrite in powers of \(x\):

\(4\sqrt{x}-\dfrac{3}{x^2}=4x^{1/2}-3x^{-2}\).

Integrate term by term:

\(\int 4x^{1/2}dx=\dfrac{8}{3}x^{3/2}\)

and

\(\int -3x^{-2}dx=3x^{-1}=\dfrac{3}{x}\).

So the answer is

\(\dfrac{8}{3}x^{3/2}+\dfrac{3}{x}+C\).

Integrate term by term:

\(\int x^3dx=\dfrac{x^4}{4}\)

and

\(\int 2x^{-1/2}dx=2\cdot\dfrac{x^{1/2}}{1/2}=4x^{1/2}\).

So the answer is

\(\dfrac{x^4}{4}+4\sqrt{x}+C\).

Integrate each term:

\(\int 12x^2dx=4x^3\)

and

\(\int -5x^{1/2}dx=-5\cdot\dfrac{x^{3/2}}{3/2}=-\dfrac{10}{3}x^{3/2}\).

So

\(Y=4x^3-\dfrac{10}{3}x^{3/2}+C\).

Integrate the gradient:

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

Use the point \((2,5)\):

\(5=2^3-4(2)+c=8-8+c\).

So \(c=5\).

Hence

\(y=x^3-4x+5\).

Integrate:

\(y=\int(x^{1/2}+2x)dx=\dfrac{2}{3}x^{3/2}+x^2+c\).

Use \((4,10)\):

\(10=\dfrac{2}{3}(4^{3/2})+4^2+c=\dfrac{2}{3}(8)+16+c\).

So

\(10=\dfrac{16}{3}+16+c=\dfrac{64}{3}+c\).

Hence \(c=-\dfrac{34}{3}\).

So the equation is

\(y=\dfrac{2}{3}x^{3/2}+x^2-\dfrac{34}{3}\).

Integrate the gradient:

\(y=\int 6x^{-2}dx=-6x^{-1}+c=-\dfrac{6}{x}+c\).

Use the point \((1,1)\):

\(1=-6+c\).

So \(c=7\).

Therefore

\(y=-\dfrac{6}{x}+7\).

First integrate:

\(\dfrac{dy}{dx}=\int 6x\,dx=3x^2+c\).

Since the gradient is 0 when \(x=1\),

\(0=3(1)^2+c\), so \(c=-3\).

Thus \(\dfrac{dy}{dx}=3x^2-3\).

Integrate again:

\(y=\int(3x^2-3)dx=x^3-3x+k\).

Use \((1,2)\):

\(2=1-3+k\), so \(k=4\).

Hence

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

The area is

\(\displaystyle \int_0^2 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^2=\dfrac{8}{3}\).

So the area is \(\dfrac{8}{3}\text{ units}^2\).

First find where the curve meets the x-axis:

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

So the roots are \(x=0\) and \(x=4\).

The area is

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

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

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

The area is

\(\displaystyle \int_0^9 x^{1/2}dx=\left[\dfrac{2}{3}x^{3/2}\right]_0^9\).

Since \(9^{3/2}=27\),

the area is \(\dfrac{2}{3}\times 27=18\).

Split the area at \(x=0\) because the curve crosses the x-axis there.

Total area

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

First part:

\(\displaystyle \int_{-1}^{0}(-x^3)dx=\frac14\).

Second part:

\(\displaystyle \int_0^2 x^3dx=\left[\dfrac{x^4}{4}\right]_0^2=4\).

Total area \(=\dfrac14+4=\dfrac{17}{4}\).

Use the volume formula

\(\displaystyle V=\pi\int y^2dx\).

Here \(y=x\), so

\(V=\pi\int_0^3 x^2dx=\pi\left[\dfrac{x^3}{3}\right]_0^3=9\pi\).

Use

\(\displaystyle V=\pi\int y^2dx\).

Since \(y=x^2\),

\(V=\pi\int_0^2 x^4dx=\pi\left[\dfrac{x^5}{5}\right]_0^2=\dfrac{32\pi}{5}\).

Use

\(\displaystyle V=\pi\int y^2dx\).

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

So

\(V=\pi\int_0^4 xdx=\pi\left[\dfrac{x^2}{2}\right]_0^4=8\pi\).

Use

\(\displaystyle V=\pi\int y^2dx\).

Here \(y=2x-x^2\), so

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

Hence

\(V=\pi\int_0^2 (4x^2-4x^3+x^4)dx\)

\(=\pi\left[\dfrac{4x^3}{3}-x^4+\dfrac{x^5}{5}\right]_0^2\)

\(=\pi\left(\dfrac{32}{3}-16+\dfrac{32}{5}\right)=\dfrac{16\pi}{15}\).