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人気のある微分積分 >

integral from 0 to 1 of sqrt(7+4t^2)

解

\int_{0}^{1} 7 + 4 t^{2} d t

解

\frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

+1

十進法表記

2.87960 \dots

解答ステップ

\int_{0}^{1} 7 + 4 t^{2} d t

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{2}{7})} \frac{7}{2} sec^{3} (u) d u

定数を除く:

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

= \frac{7}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{7})} sec^{3} (u) d u

積分換算を適用する

= \frac{7}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{7})} sec (u) d u

\int_{0}^{a r c t a n (\frac{2}{7})} sec (u) d u = ln (\frac{2}{7} + \frac{11}{7})

= \frac{7}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7})

簡素化

\frac{7}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7}) : \frac{1}{4} 14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7})

= \frac{1}{4} 14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7})

限度を計算する:

\frac{11}{7}

= \frac{1}{4} (14 \cdot \frac{11}{7} + 7 ln (\frac{2}{7} + \frac{11}{7}))

簡素化

= \frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

グラフ

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