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人気のある微分積分 >

integral from 0 to 3 of m^2(4m^2+2)^3

解

\int_{0}^{3} m^{2} (4 m^{2} + 2)^{3} d m

解

42 (- \frac{181764 2}{35} + 267392)

+1

十進法表記

172365.94285 \dots

解答ステップ

\int_{0}^{3} m^{2} (4 m^{2} + 2)^{3} d m

置換積分法を適用する

= \int_{0}^{32} \frac{4 u ^{2} ( u ^{2} + 1 ) ^{3}}{2} d u

定数を除く:

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

= \frac{4}{2} \cdot \int_{0}^{32} u^{2} (u^{2} + 1)^{3} d u

\frac{4}{2} = 22

= 22 \cdot \int_{0}^{32} u^{2} (u^{2} + 1)^{3} d u

部分積分を適用する

= 22 [\frac{1}{3} (u^{3} (u^{2} + 1)^{3} - 3 \cdot \int 2 u^{4} (u^{2} + 1)^{2} d u)]_{0}^{32}

\int 2 u^{4} (u^{2} + 1)^{2} d u = 2 (\frac{u ^{9}}{9} + \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5})

= 22 [\frac{1}{3} (u^{3} (u^{2} + 1)^{3} - 3 \cdot 2 (\frac{u ^{9}}{9} + \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}))]_{0}^{32}

簡素化

22 [\frac{1}{3} (u^{3} (u^{2} + 1)^{3} - 3 \cdot 2 (\frac{u ^{9}}{9} + \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}))]_{0}^{32} : 22 [\frac{1}{3} (u^{3} (u^{2} + 1)^{3} - 6 (\frac{u ^{9}}{9} + \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}))]_{0}^{32}

= 22 [\frac{1}{3} (u^{3} (u^{2} + 1)^{3} - 6 (\frac{u ^{9}}{9} + \frac{2 u ^{7}}{7} + \frac{u ^{5}}{5}))]_{0}^{32}

限度を計算する:

2 (- \frac{181764 2}{35} + 267392)

= 22 \cdot 2 (- \frac{181764 2}{35} + 267392)

簡素化

= 42 (- \frac{181764 2}{35} + 267392)

グラフ

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