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人気のある微分積分 >

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

解

\int_{0}^{π} 2 x sin (2 x^{2}) d x

解

\frac{1}{2} (- cos (2 π^{2}) + 1)

+1

十進法表記

0.18515 \dots

解答ステップ

\int_{0}^{π} 2 x sin (2 x^{2}) 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}^{π} x sin (2 x^{2}) d x

u置換積分法を適用する

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

共通積分を使用する:

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

= 2 \cdot \frac{1}{4} [- cos (u)]_{0}^{2 π^{2}}

簡素化

2 \cdot \frac{1}{4} [- cos (u)]_{0}^{2 π^{2}} : \frac{1}{2} [- cos (u)]_{0}^{2 π^{2}}

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

限度を計算する:

- cos (2 π^{2}) + 1

= \frac{1}{2} (- cos (2 π^{2}) + 1)

グラフ

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