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#