Evaluate? int_pi^(2pi)theta d theta2ππθdθ

1 Answer
Dec 14, 2016

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The integral int_pi^(2pi)theta d theta2ππθdθ, by its very definition represents the area under the graph of the integrand y=thetay=θ (or y=xy=x if you prefer) from theta=piθ=π to theta=2piθ=2π (or x=pix=π to x=2pix=2π)

So we have a trapezium of width piπ and lengths piπ and 2pi2π
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.