$\displaystyle For\ half\ range\ cosine\ series\ b_n=0 \\[2ex] \displaystyle Here\ l=2 \\[2ex] \displaystyle a_0=\frac{2}{l}\int_0^lf\left(x\right)dx \\[2ex] \displaystyle =\frac{2}{2}\int_0^2(2x-x^{2\ })dx \\[2ex] \displaystyle ={\left[2\bullet{}\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^2 \\[2ex] \displaystyle =\left(2^2-\frac{x^3}{3}\right)-\left(0-0\right) \\[2ex] \displaystyle =\frac{4}{3} \\[2ex] $

$\displaystyle a_n=\frac{2}{l}\int_0^lf\left(x\right)\cos{\frac{n\pi{}x}{l}dx} \\[2ex] \displaystyle =\frac{2}{l}\int_0^l(2x-x^{2\ })\cos{\frac{n\pi{}x}{l}dx} \\[2ex] \displaystyle ={\left[\left(2x-x^{2\ }\right)\sin{\frac{n\pi{}x}{2}}\bullet{}\frac{2}{n\pi{}}-\left(2-2x\right)\bullet{}-cos\frac{n\pi{}x}{2}\bullet{}{\left(\frac{2}{n\pi{}}\right)}^2+\left(-2\right)\bullet{}-sin\frac{n\pi{}x}{2}\bullet{}{\left(\frac{2}{n\pi{}}\right)}^3\right]}_0^2 \\[2ex] \displaystyle =\left[0+(-2)\bullet{}\cos{n\pi{}\bullet{}\frac{4}{n^2{\pi{}}^2}+0}\right]-\left[0+\left(2\right)\bullet{}1\bullet{}\frac{4}{n^2{\pi{}}^2}+0\right] \\[2ex] \displaystyle =\frac{8\left[-{\left(-1\right)}^n-1\right]}{n^2{\pi{}}^2} \\[2ex] \displaystyle \therefore{}Half\ range\ fourier\ cosine\ series\ is \\[2ex] $

$\displaystyle f(x)=\frac{a0}{2}+\sum_{n=1}^{\infty{}}an\ \cos{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle \therefore{}2x-x^{2\ }=\frac{2}{3}+\sum_{n=1}^{\infty{}}\frac{8\left[-{\left(-1\right)}^n-1\right]}{n^2{\pi{}}^2}\cos{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle =\frac{2}{3}-\frac{8}{{\pi{}}^2}\left[0+\frac{2}{2^2}\cos{\frac{2\pi{}x}{2}+0+\frac{2}{4^2}\cos{\frac{4\pi{}x}{2}}}+…\right] \\[2ex] \displaystyle =\frac{2}{3}-\frac{8}{{\pi{}}^2}\times{}\frac{2}{2^2}\left[\cos{\frac{\pi{}x}{1^2}+\cos{\frac{2\pi{}x}{2^2}}}+…\right] \\[2ex] \displaystyle =\frac{2}{3}-\frac{4}{{\pi{}}^2}\left[\cos{\frac{\pi{}x}{1^2}+\cos{\frac{2\pi{}x}{2^2}}}+\cos{\frac{3\pi{}x}{3^2}}+…\right] \\[2ex]$