Use the Riemann integral.
Area under #f(x)# from #[a, b]# is #int_a^bf(x) dx#
So, the area under #r(theta)=(sintheta)/theta-theta^3-theta# where #theta in [pi/12, pi/3]# is #int_(pi/12)^(pi/3)r(theta) \quad d theta#
Let's integrate the indefinite and ignore the constant:
#int \quad r(theta) \quad d theta#
#=int\quad (sintheta)/theta-theta^3-theta \quad d theta#
#=int\quad (sintheta)/theta \quad d theta - int\quad theta^3 \quad d theta - int\quad theta \quad d theta#
Now, #int\quadsintheta/theta\quadd theta="Si"(theta)+C# which is a very interesting function. We have entered the lair of Trigonometric Integrals. There is no closed form of #"Si"#, so we would have to deal with #"Si"(pi/3)-"Si"(pi/12)~~0.724654...#
Anyways, we now integrate the (trivial) rest, again, ignoring the constant
#int\quad theta^3 \quad d theta=(theta^4)/4#
#int\quad theta \quad d theta=(theta^2)/2#
Finally, we piece everything together:
#int_(pi/12)^(pi/3)r(theta) \quad d theta=|["Si"(theta)-((theta^2)(theta^2+2))/4]|_(pi/12)^(pi/3)#
#~~0.724564...-((pi/3)^2((pi/3)^2+2))/4+((pi/12)^2((pi/12)^2+2))/4#
#~~0.724564...-0.813512...#
#~~-0.08885...#
#:.# The (unsigned) area under #r(theta)# is approx. #0.0889#
