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人気のある微分積分 >

integral from 0 to 1 of tsin(t)

解

\int_{0}^{1} t sin (t) d t

解

sin (1) - cos (1)

+1

十進法表記

0.30116 \dots

解答ステップ

\int_{0}^{1} t sin (t) d t

部分積分を適用する

= [- t cos (t) - \int - cos (t) d t]_{0}^{1}

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

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

簡素化

= [- t cos (t) + sin (t)]_{0}^{1}

限度を計算する:

sin (1) - cos (1)

= sin (1) - cos (1)

グラフ

人気の例

integral from 0 to 30 of x/(sqrt(4+2x))

\int_{0}^{30} \frac{x}{4 + 2 x} d x

integral from 1 to 3 of (x^2+9)

\int_{1}^{3} (x^{2} + 9) d x

integral from-2 to 2 of 4xsqrt(4-x^2)

\int_{- 2}^{2} 4 x 4 - x^{2} d x

integral from 0 to x of arcsin(t)

\int_{0}^{x} arcsin (t) d t

integral from 2 to 3 of-3/(x^2)

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