integral from 0 to 2 of xcos((npi)/2 x)

解

\int_{0}^{2} x cos (\frac{nπ}{2} x) d x

\frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

解答ステップ

\int_{0}^{2} x cos (\frac{nπ}{2} x) d x

\frac{nπ}{2} = nπ \frac{1}{2}

= \int_{0}^{2} x cos (nπ \frac{1}{2} x) d x

u置換積分法を適用する

= \int_{0}^{πn} \frac{4 u cos ( u )}{π ^{2} n ^{2}} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} u cos (u) d u

部分積分を適用する

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{πn}

\int sin (u) d u = - cos (u)

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn}

簡素化

= \frac{4}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

限度を計算する:

(- 1)^{n} - 1

= \frac{4}{π ^{2} n ^{2}} ((- 1)^{n} - 1)

簡素化

= \frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}