To solve this, first factor out one cosine and write the rest in terms of sine:

\(\displaystyle \int \sin^{6}x \, \cos^{3}x\,dx=\int \sin^{6}x \, \cos^{2}x\,\cos x\, dx=\int \sin^{6}x (1-\sin^{2}x)\cos x\, dx\)

From here, we can do a substitution where u=sin(x) and du=cos(x)dx:

\(\displaystyle \int \sin^{6}x (1-\sin^{2}x)\cos x\, dx=\int u^{6}(1-u^{2})du=\int(u^{6}-u^{8})du=\frac{u^{7}}{7}-\frac{u^{9}}{9}+C\)

Finally, re-substitute sin(x) for u:

\(\displaystyle \frac{u^{7}}{7}-\frac{u^{9}}{9}+C=\frac{\sin^{7}x}{7}-\frac{\sin^{9}x}{9}+C\)

The formula for volume of the region revolved around the x-axis is given as 

\(\displaystyle V=\pi \int_{a}^{b} r^2\, dx\)

where \(\displaystyle r=f(x)\)

As such

\(\displaystyle V=\pi \int_{0}^{10} (-x^2+10x+3)^2\,dx = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx\)

When taking the integral, we use the inverse power rule which states

\(\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1}\)

Applying this rule term by term we get

\(\displaystyle = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx = \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}\)

And by the corollary of the Fundamental Theorem of Calculus 

\(\displaystyle \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}=\pi \left ( \left [ \frac{(10)^5}{5}-\frac{20(10)^4}{4}+\frac{94(10)^3}{3}+\frac{60(10)^2}{2}+9(10) \right ]-0 \right )\)

\(\displaystyle =\frac{13270}{3}\pi\)

As such the volume is

\(\displaystyle \frac{13270}{3}\pi\) units cubed

Since we are revolving a function of \(\displaystyle x\) around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

\(\displaystyle \int_{0}^{4}2\pi x (\sqrt{x})dx\)

\(\displaystyle =\int_{0}^{4}2\pi x^{3/2}dx\)

\(\displaystyle =[2\pi x^{5/2}]^{4}_{0}\)

\(\displaystyle =\frac{128}{5}\pi\).

Find the volume of the solid region swept out by the area bounded by the the functions \(\displaystyle \small \small \small y = \sqrt{3x}\)  and \(\displaystyle \small \small \small y = x^2\) about the \(\displaystyle \small x\)-axis. 

The cross sectional area of the solid can be written as a function of \(\displaystyle \small x\) by first finding the cross-sectional areas swept out by each of the individual functions as they rotate about the x-axis, and then subtracting the inner area. 

Visualize a line drawn perpendicularly to the x-axis at some point meeting the function \(\displaystyle \small \small \small y = \sqrt{3x}\). As we rotate about the x-axis, we sweep out a circular cross-sectional area with a radius equal to the function value \(\displaystyle \small \small \small y = \sqrt{3x}\) at the corresponding value of \(\displaystyle \small x\). The area of this disk is therefore:

\(\displaystyle \small \small \small \small A_1(x)=\pi (\sqrt{3x})^2=\pi(3x)=3\pi x\)

Similarly for the function \(\displaystyle \small \small \small y = x^2\), the cross-sectional area is: 

\(\displaystyle \small \small A_2(x)=\pi (x^2)^2=\pi x^4\)

The plot of the two functions helps visualize the geometry. The parabola is clearly the "inner" radius, and therefore we must subtract the areas \(\displaystyle \small A_2(x)\) from \(\displaystyle \small A_1(x)\) to find the cross-sectional area of the solid:

\(\displaystyle \small \small A(x)=A_1(x)-A_2(x)\)

 \(\displaystyle \small \small A(x)=3\pi x-\pi x^4\)

Now we must find the limits of integration by finding the intersection points. 

\(\displaystyle \small \sqrt{3x}=x^2\)

Solve for \(\displaystyle \small x\)

\(\displaystyle \small \small 3x=x^4\: \: \: \rightarrow\: \: \:3=x^3 \: \: \:\rightarrow\: \: \: x=\sqrt[3]{3}\)

The other solution is \(\displaystyle \small x = 0\). Now simply integrate the cross sectional area \(\displaystyle \small A(x)\)over the interval \(\displaystyle \small [0,\sqrt[3]{3} ]\) to find the volume of the solid, 

\(\displaystyle \small V=\small \int_0^{\sqrt[3]3} (3\pi x-\pi x^4)dx\)

\(\displaystyle \small \small \small \small \small \small V=\left[\frac{3\pi}{2} x^2 - \frac{\pi}{5} x^5\right ]_{0}^{\sqrt[3]{3}}\: \: \:\)

\(\displaystyle \small V=\frac{3}{2}\pi (\sqrt[3]{3})^2 - \frac{1}{5}\pi (\sqrt[3]{3})^5\)

 To simplify further, write the terms with a common denominator and factor. 

\(\displaystyle \small V=\frac{15}{10}\pi (\sqrt[3]{3})^2 - \frac{2}{10}\pi (\sqrt[3]{3})^5\)

\(\displaystyle \small V=\frac{15}{10}\pi (\sqrt[3]{3})^2 - \frac{2}{10}\pi (\sqrt[3]{3})^5\)

\(\displaystyle V=\frac{1}{10}\pi(\sqrt[3]{3})^2 \left[15-2(\sqrt[3]{3})^3\right]\)

\(\displaystyle V=\frac{1}{10}\pi(\sqrt[3]{3})^2 \left[15-6\right]\)

\(\displaystyle V=\frac{9}{10}\pi(\sqrt[3]{3})^2\)

The volume of a solid of revolution for a given function is found using the disk method. Answer is obtained upon evaluating the following integral:

\(\displaystyle \\\int_{0}^{2}\pi(x^2+1)^2\:dx\\\\=\pi\int_{0}^{2}x^4+2x^2+1\:dx\\\\=\left.\begin{matrix} \pi(\frac{x^5}{5}+\frac{2x^3}{3}+x) \end{matrix}\right|_{0}^{2}\\\\=\frac{206\pi}{15}\)

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f (x) = \ln (x + 1)\), so

\(\displaystyle f'(x) = \frac{1}{x+1}\), and the integral to be approximated is

\(\displaystyle \int_{0}^{2} \sqrt{1 + \left ( \frac{1}{x+1} \right ) ^{2}} \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{2} \sqrt{\frac{x^{2}+2x+1}{x^{2}+2x+1} + \frac{1}{x^{2}+2x+1} } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{2} \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1} } \; \mathrm{d}x\)

We divide the interval \(\displaystyle [0,2]\)  into four subintervals of width \(\displaystyle \Delta x = 2 \div 4 = 0.5\) each, so 

\(\displaystyle x_{0} = 0, x_{1} = 0.5, x_{2} = 1, x_{3} = 1.5, x_{4} = 2\)

By Simpson's rule, we can estimate the integral by evaluating

\(\displaystyle \frac{\Delta x}{3} \left ( g (x_{0}) + 4g(x_{1}) + 2g(x_{2}) + 4 g (x_{3}) + g (x_{4}) \right )\)

\(\displaystyle = \frac{0.5 }{3} \left ( g (0) + 4g(0.5) + 2g(1) + 4 g (1.5) + g (2) \right )\)

where \(\displaystyle g(x) = \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1}\)

We evaluate \(\displaystyle g\) at these points, then substitute:

\(\displaystyle g(0) = \sqrt{\frac{0^{2}+2 \cdot 0+2}{0^{2}+2 \cdot 0+1} } \approx 1.4142\)

\(\displaystyle g(0.5) = \sqrt{\frac{0.5^{2}+2 \cdot 0.5+2}{0.5^{2}+2 \cdot 0.5+1} } \approx 1.2019\)

\(\displaystyle g(1) = \sqrt{\frac{1^{2}+2 \cdot 1+2}{1^{2}+2 \cdot 1+1} } \approx 1.1180\)

\(\displaystyle g(1.5) = \sqrt{\frac{1.5^{2}+2 \cdot 1.5+2}{1.5^{2}+2 \cdot 1.5+1} } \approx 1.0770\)

\(\displaystyle g(2) = \sqrt{\frac{2^{2}+2 \cdot 2+2}{2^{2}+2 \cdot 2+1} } \approx 1.0541\)

The approximation is therefore

\(\displaystyle \approx \frac{0.5 }{3} \left ( 1.4142 +4 \cdot 1.2019 + 2\cdot 1.1180 +4 \cdot 1.0770 +1.0541 \right )\)

\(\displaystyle \approx \frac{0.5 }{3} \left ( 1.4142 +4.8076 + 2.2360 +4.3080 +1.0541 \right )\)

\(\displaystyle \approx \frac{0.5 }{3} \left (13.8199 \right ) \approx 2.303\)

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}\)

so

\(\displaystyle f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}\) .

The above integral becomes 

\(\displaystyle \int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x\)

\(\displaystyle \int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x\)

Substitute \(\displaystyle u = x + 1\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 1\), \(\displaystyle \mathrm{d} u = \mathrm{d} x\), and the integral becomes

\(\displaystyle \int_{1}^{4} \sqrt{u} \; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} \; \mathrm{d}u\)

\(\displaystyle = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}\)

\(\displaystyle = \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}\)

\(\displaystyle = \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}\)

\(\displaystyle = \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}\)

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )\), so 

\(\displaystyle f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x\)

The integral becomes

\(\displaystyle \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x\)

Use substitution - set \(\displaystyle u = 2x\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 2\), and \(\displaystyle \mathrm{d} x = \frac{1}{2}\mathrm{d} u\). The bounds of integration become \(\displaystyle 2 \cdot 0 = 0\) and \(\displaystyle 2 \cdot \frac{\pi}{8} = \frac{\pi}{4}\), and the integral becomes

\(\displaystyle \frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u\)

\(\displaystyle =\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}\)

\(\displaystyle =\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |\)

\(\displaystyle =\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |\)

\(\displaystyle =\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1\)

\(\displaystyle =\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0\)

\(\displaystyle =\frac{1}{2} \ln \left (1 + \sqrt{2} \right )\)

The length of the curve of \(\displaystyle f(x)\) on the interval \(\displaystyle [a,b]\) can be determined by evaluating the integral

\(\displaystyle \int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x\).

\(\displaystyle f(x) = \frac{1}{3} \left ( \cosh 3 x \right )\), so

\(\displaystyle f'(x) = \frac{1}{3} \cdot 3 \sinh 3 x = \sinh 3 x\)

and the integral becomes

\(\displaystyle \int_{0}^{2} \sqrt{1 + \sinh^{2} 3 x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{2} \sqrt{ \cosh^{2} 3 x } \; \mathrm{d}x\)

\(\displaystyle = \int_{0}^{2} \cosh 3 x \; \mathrm{d}x\)

Use substitution - set \(\displaystyle u =3x\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 3\), and \(\displaystyle \mathrm{d} x = \frac{1}{3}\mathrm{d} u\). The bounds of integration become 0 and 6; the integral becomes

\(\displaystyle = \int_{0}^{6} \frac{1}{3} \cosh u \; \mathrm{d}u\)

\(\displaystyle =\frac{1}{3} \int_{0}^{6} \cosh u \; \mathrm{d}u\)

\(\displaystyle = \frac{1}{3} \cdot \left.\begin{matrix} \sinh u\\ \\ \end{matrix}\right|\left.\begin{matrix} 6\\ \\ 0 \end{matrix}\right\)

\(\displaystyle =\frac{1}{3}\left ( \sinh 6 - \sinh 0 \right ) = \frac{1}{3} \sinh 6 =\frac{1}{3 } \cdot \frac{ e^{6} - e^{-6}}{2} = \frac{ \left (e^{6} - e^{-6} \right )e^{6}}{6e^{6}} = \frac{ e^{12} - 1 }{6e^{6}}\)

The average value of the function \(\displaystyle f(x) = \tan x\) on the interval \(\displaystyle \left [\frac{ \pi }{6} , \frac{ \pi }{3}\right ]\) is equal to 

\(\displaystyle \frac{1}{ \frac{ \pi}{3}-( \frac{ \pi}{6})} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x\)

\(\displaystyle = \frac{1}{ \frac{ \pi}{6}} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x\)

\(\displaystyle = \frac{6}{ \pi} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x\)

\(\displaystyle = \frac{6}{ \pi} \cdot \left.\begin{matrix} \ln\left | \sec x \right |\\ \\ \end{matrix} \right | \begin{matrix} \frac{\pi}{3}\\ \\ \frac{ \pi}{6} \end{matrix}\)

\(\displaystyle = \frac{6}{ \pi} \left ( \ln\left | \sec \frac{ \pi}{3} \right | - \ln\left | \sec \frac{ \pi}{6} \right | \right )\)

\(\displaystyle = \frac{6}{ \pi} \left ( \ln 2 - \ln \frac{2\sqrt{3}}{3} \right )\)

\(\displaystyle = \frac{6}{ \pi} \ln \left (2 \div \frac{2\sqrt{3}}{3} \right )\)

\(\displaystyle = \frac{6}{ \pi} \ln \sqrt{3}= \frac{6}{ \pi} \cdot \frac{1}{2} \cdot \ln 3 = \frac{3 \ln 3}{\pi}\)