integral from 0 to 1 of 2xcos(npix)

解

\int_{0}^{1} 2 x cos (nπ x) d x

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

解答ステップ

\int_{0}^{1} 2 x cos (nπ x) d x

定数を除く:

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

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

u置換積分法を適用する

= 2 \cdot \int_{0}^{πn} \frac{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

= 2 \cdot \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} u cos (u) d u

部分積分を適用する

= 2 \cdot \frac{1}{π ^{2} n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{πn}

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

= 2 \cdot \frac{1}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn}

簡素化

2 \cdot \frac{1}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn} : \frac{2}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

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

限度を計算する:

(- 1)^{n} - 1

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

簡素化

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