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人気のある微分積分 >

integral from 1 to 2 of cos(sqrt(x-1))

解

\int_{1}^{2} cos (x - 1) d x

解

2 (sin (1) + cos (1) - 1)

+1

十進法表記

0.76354 \dots

解答ステップ

\int_{1}^{2} cos (x - 1) d x

u置換積分法を適用する

= \int_{0}^{1} cos (u) d u

u置換積分法を適用する

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

定数を除く:

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

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

部分積分を適用する

= 2 [v sin (v) - \int sin (v) d v]_{0}^{1}

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

= 2 [v sin (v) - (- cos (v))]_{0}^{1}

簡素化

= 2 [v sin (v) + cos (v)]_{0}^{1}

限度を計算する:

sin (1) + cos (1) - 1

= 2 (sin (1) + cos (1) - 1)

グラフ

人気の例

integral from 1 to e^7 of ((ln(x))^2)/x

\int_{1}^{e^{7}} \frac{( ln ( x ) ) ^{2}}{x} d x

integral from 1 to 2 of 1/(sqrt(2x-x^2))

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

integral from 1 to 4 of sqrt(1+(2x)^2)

\int_{1}^{4} 1 + (2 x)^{2} d x

integral from 0 to 1 of 1/(xe^{x^2)}

\int_{0}^{1} \frac{1}{x e ^{x^{2}}} d x

integral from 0 to pi of 1 (sin(x))

\int_{0}^{π} \frac{d}{1} (sin (x)) d x