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人気のある微分積分 >

integral from 0 to 1 of 6x^2cos(x^3)

解

\int_{0}^{1} 6 x^{2} cos (x^{3}) d x

解

2 sin (1)

+1

十進法表記

1.68294 \dots

解答ステップ

\int_{0}^{1} 6 x^{2} cos (x^{3}) d x

定数を除く:

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

= 6 \cdot \int_{0}^{1} x^{2} cos (x^{3}) d x

u置換積分法を適用する

= 6 \cdot \int_{0}^{1} \frac{cos ( u )}{3} d u

定数を除く:

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

= 6 \cdot \frac{1}{3} \cdot \int_{0}^{1} cos (u) d u

共通積分を使用する:

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

= 6 \cdot \frac{1}{3} [sin (u)]_{0}^{1}

簡素化

6 \cdot \frac{1}{3} [sin (u)]_{0}^{1} : 2 [sin (u)]_{0}^{1}

= 2 [sin (u)]_{0}^{1}

限度を計算する:

sin (1)

= 2 sin (1)

グラフ

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