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人気のある微分積分 >

integral from 0 to 2 of sqrt(5+4x^2)

解

\int_{0}^{2} 5 + 4 x^{2} d x

解

21 + \frac{5}{4} ln (\frac{4}{5} + \frac{21}{5})

+1

十進法表記

6.26384 \dots

解答ステップ

\int_{0}^{2} 5 + 4 x^{2} d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{4}{5})} \frac{5}{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{5}{2} \cdot \int_{0}^{a r c t a n (\frac{4}{5})} sec^{3} (u) d u

積分換算を適用する

= \frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{4}{5})} sec (u) d u

\int_{0}^{a r c t a n (\frac{4}{5})} sec (u) d u = ln (\frac{4}{5} + \frac{21}{5})

= \frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} ln (\frac{4}{5} + \frac{21}{5})

簡素化

\frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} ln (\frac{4}{5} + \frac{21}{5}) : \frac{1}{4} 10 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{4}{5})} + 5 ln (\frac{4}{5} + \frac{21}{5})

= \frac{1}{4} 10 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{4}{5})} + 5 ln (\frac{4}{5} + \frac{21}{5})

限度を計算する:

\frac{2 21}{5}

= \frac{1}{4} (10 \cdot \frac{2 21}{5} + 5 ln (\frac{4}{5} + \frac{21}{5}))

簡素化

= 21 + \frac{5}{4} ln (\frac{4}{5} + \frac{21}{5})

グラフ

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