Evaluate? int_pi^(2pi)theta d theta∫2ππθdθ
1 Answer
Dec 14, 2016
The integral
So we have a trapezium of width
So the area of that trapezium is given by:
A = 1/2(a+b)hA=12(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.