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人気のある微分積分 >

integral from 0 to t of usin(u)

解

\int_{0}^{t} u sin (u) d u

解

sin (t) - t cos (t)

解答ステップ

\int_{0}^{t} u sin (u) d u

部分積分を適用する

= [- u cos (u) - \int - cos (u) d u]_{0}^{t}

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

= [- u cos (u) - (- sin (u))]_{0}^{t}

簡素化

= [- u cos (u) + sin (u)]_{0}^{t}

限度を計算する:

sin (t) - t cos (t)

= sin (t) - t cos (t)

人気の例

integral from 4 to 7 of-9log_{3}(t)

\int_{4}^{7} - 9 lo g_{3} (t) d t

integral from 1 to pi+1 of sin(2x)

\int_{1}^{π + 1} sin (2 x) d x

integral from 1 to 4 of 6x

\int_{1}^{4} 6 x d x

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

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

integral from-2 to 1 of (x^2+x-2)

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