Science:Math Exam Resources/Courses/MATH307/December 2012/Question 04 (e)/Solution 1

Using the hint given in the question, we have

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\cos (2\pi t)(t^2-t)dt& =\underset{0}{\overset{1}{\int }}\frac{1}{2}(e^{2\pi it}+e^{-2\pi it})(t^2-t)dt\\ & =\underset{0}{\overset{1}{\int }}\frac{1}{2}{e}^{2\pi it}(t^2-t)dt+\underset{0}{\overset{1}{\int }}\frac{1}{2}{e}^{-2\pi it}(t^2-t)dt\\ & =\frac{1}{2}\underset{0}{\overset{1}{\int }}(t^2-t)e^{2\pi it}dt+\frac{1}{2}\underset{0}{\overset{1}{\int }}(t^2-t)e^{-2\pi it}dt\\ & =\frac{1}{2}⟨t^2-t,e^{-2\pi it}⟩+\frac{1}{2}⟨t^2-t,e^{2\pi it}⟩\\ & =\frac{1}{2}{c}_{-1}+\frac{1}{2}{c}_1\end{array}$

From part (d) we know that $c_n=\frac{1}{2{\pi }^2{n}^2}$, which we plug into our equation above to calculate

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\cos (2\pi t)(t^2-t)dt& =\frac{1}{2}\left(\frac{1}{2{\pi }^2(-1{)}^2}\right)+\frac{1}{2}\left(\frac{1}{2{\pi }^2(1{)}^2}\right)\\ & =\frac{1}{4{\pi }^2}+\frac{1}{4{\pi }^2}=\frac{1}{2{\pi }^2}\end{array}$