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人気のある微分積分 >

integral from 0 to l of lsin((npix)/l)

解

\int_{0}^{l} l sin (\frac{nπ x}{l}) d x

解

\frac{l ^{2} ( - ( - 1 ) ^{n} + 1 )}{πn}

解答ステップ

\int_{0}^{l} l sin (\frac{nπ x}{l}) d x

定数を除く:

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

= l \cdot \int_{0}^{l} sin (\frac{nπ x}{l}) d x

u置換積分法を適用する

= l \cdot \int_{0}^{nπ l} \frac{sin ( \frac{u}{l} )}{πn} d u

定数を除く:

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

= l \frac{1}{πn} \cdot \int_{0}^{nπ l} sin (\frac{u}{l}) d u

u置換積分法を適用する

= l \frac{1}{πn} \cdot \int_{0}^{πn} sin (v) l d v

定数を除く:

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

= l \frac{1}{πn} l \cdot \int_{0}^{πn} sin (v) d v

共通積分を使用する:

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

= l \frac{1}{πn} l [- cos (v)]_{0}^{πn}

簡素化

l \frac{1}{πn} l [- cos (v)]_{0}^{πn} : l^{2} \frac{1}{πn} [- cos (v)]_{0}^{πn}

= l^{2} \frac{1}{πn} [- cos (v)]_{0}^{πn}

限度を計算する:

- (- 1)^{n} + 1

= l^{2} \frac{1}{πn} (- (- 1)^{n} + 1)

簡素化

= \frac{l ^{2} ( - ( - 1 ) ^{n} + 1 )}{πn}

人気の例

integral from 0 to 2 of xsin(pix)

\int_{0}^{2} x sin (π x) d x

integral from 0 to 1 of (1+e^x)/(x+e^x)

\int_{0}^{1} \frac{1 + e ^{x}}{x + e ^{x}} d x

integral from-11 to 11 of |10x|-(x^2-11)

\int_{- 11}^{11} ∣ 10 x ∣ - (x^{2} - 11) d x

integral from 0 to a of sqrt(a^2-r^2)r

\int_{0}^{a} a^{2} - r^{2} r d r

integral from 2 to 6 of pi(2/y)^2

\int_{2}^{6} π (\frac{2}{y})^{2} d y