Evaluate? #int_pi^(2pi)theta d theta#

1 Answer
Dec 14, 2016

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The integral #int_pi^(2pi)theta d theta#, by its very definition represents the area under the graph of the integrand #y=theta# (or #y=x# if you prefer) from #theta=pi# to #theta=2pi# (or #x=pi# to #x=2pi#)

So we have a trapezium of width #pi# and lengths #pi# and #2pi#
So the area of that trapezium is given by:

# A = 1/2(a+b)h#
# \ \ \ = 1/2(pi+2pi)pi#
# \ \ \ = 1/2(3pi)pi#
# \ \ \ = (3pi^2)/2 #QED

We can also evaluate the integral using calculus to confirm the result:

#int_pi^(2pi)theta d theta = [1/2theta^2]_pi^(2pi)#
# " " = 1/2[theta^2]_pi^(2pi)#
# " " = 1/2((2pi)^2-(pi)^2)#
# " " = 1/2(4pi^2-pi^2)#
# " " = 1/2(3pi^2)#
# " " = (3pi^2)/2#, reassuringly confirming our result.