Answer
$\displaystyle \frac{1}{2π} - \frac{1}{\pi^2} ≈ 0.057833$
Work Step by Step
Integration by parts formula: $\displaystyle \int fg' $ $\displaystyle dx = fg-\int f'g$ $dx$
$\displaystyle \int_0^{1/2} x\cos(\pi x)dx$
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Let $\displaystyle f = x$ and $\displaystyle g' = \cos(πx)$
thus $\displaystyle f' = 1$ and $\displaystyle g = \frac{1}{π}\sin(\pi x)$
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$\displaystyle \int_0^{1/2} x\cos(\pi x)dx = (x[\frac{1}{\pi}\sin(\pi x)])|_0^{1/2} - \int_0^{1/2}(1)(\frac{1}{π}\sin(\pi x))dx$
$\displaystyle [\frac{x}{\pi}\sin(\pi x)]_0^{1/2} - [-\frac{1}{π^2}\cos(πx))]_0^{1/2}$
$\displaystyle [\frac{\frac{1}{2}}{π}\sin(\frac{π}{2}) - \frac{0}{\pi}\sin(0)] - [-\frac{1}{π^2}\cos(\frac{π}{2})-(-\frac{1}{π^2}\cos(0))]$
$\displaystyle [\frac{1}{2π}(1) - 0] - [-\frac{1}{π^2}(0) + \frac{1}{\pi^2}(1)]$
$\displaystyle \frac{1}{2π} - \frac{1}{\pi^2} ≈ 0.057833$
