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人気のある微分積分 >

integral from 0 to sqrt(pi of)-xsin(x^2)

解

\int_{0}^{π} - x sin (x^{2}) d x

解

- 1

解答ステップ

\int_{0}^{π} - x sin (x^{2}) d x

定数を除く:

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

= - \int_{0}^{π} x sin (x^{2}) d x

u置換積分法を適用する

= - \int_{0}^{π} \frac{sin ( u )}{2} d u

定数を除く:

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

= - \frac{1}{2} \cdot \int_{0}^{π} sin (u) d u

共通積分を使用する:

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

= - \frac{1}{2} [- cos (u)]_{0}^{π}

限度を計算する:

2

= - \frac{1}{2} \cdot 2

簡素化

= - 1

グラフ

人気の例

integral from 9 to 15 of x

\int_{9}^{15} x d x

integral from 2 to 4 of k[1-(x-3)^2]

\int_{2}^{4} k [1 - (x - 3)^{2}] d x

integral from 0 to pi of e^xcos(nx)

\int_{0}^{π} e^{x} cos (n x) d x

integral from 2 to 3 of x-1

\int_{2}^{3} x - 1 d x

integral from pi to 0 of sin^2(x)

\int_{π}^{0} sin^{2} (x) d x