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人気のある微分積分 >

integral from 1 to 2 of xcos(x)

解

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

解

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

+1

十進法表記

0.02067 \dots

解答ステップ

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

部分積分を適用する

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

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

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

簡素化

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

限度を計算する:

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

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

グラフ

人気の例

integral from 0 to 4 of 5e^x

\int_{0}^{4} 5 e^{x} d x

integral from-pi/2 to pi/2 of cos^2(t)

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (t) d t

integral from 0 to 2 of pi(x^2+4)^2

\int_{0}^{2} π (x^{2} + 4)^{2} d x

integral from 0 to 1 of 1/(1+t)

\int_{0}^{1} \frac{1}{1 + t} d t

integral from 0 to 0.19 of (sin(x))/x

\int_{0}^{0.19} \frac{sin ( x )}{x} d x