Find the surface area generated from x = 1 to x = 2:
[tex]y=\frac{x^3}{12}+\frac{1}{x}[/tex]
The definite integral of the above function from x = 1 to x = 2, will be used to generate the surface area
[tex]\begin{gathered} \int ^2_1\frac{x^3}{12}+\frac{1}{x}dx \\ \mathrm{Apply\: the\: Sum\: Rule}\colon\quad \int f\mleft(x\mright)\pm g\mleft(x\mright)dx=\int f\mleft(x\mright)dx\pm\int g\mleft(x\mright)dx \end{gathered}[/tex][tex]\begin{gathered} =\int ^2_1\frac{x^3}{12}dx+\int ^2_1\frac{1}{x}dx \\ =\int ^2_1\lbrack\frac{x^4}{4(12)}+\ln x\rbrack \end{gathered}[/tex][tex]\begin{gathered} =\int ^2_1\lbrack\frac{x^4}{4(12)}+\ln x\rbrack \\ =\int ^2_1\lbrack\frac{x^4}{4(12)}+\int ^2_1\ln x\rbrack \\ =(\frac{2^4}{48}-\frac{1^4}{48})+(\ln 2-\ln 1) \\ =(\frac{16}{48}-\frac{1}{48})+\ln 2-0 \\ =\frac{15}{48}+\ln (2) \\ \frac{5}{16}+\ln (2) \end{gathered}[/tex]
Hecne the surface area of the function = 5/16 + In(2)
[tex]\frac{5}{16}+\ln (2)[/tex]
