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人気のある微分積分 >

integral from 0 to 1 of sqrt(35+x^2)

解

\int_{0}^{1} 35 + x^{2} d x

解

3 + \frac{35}{4} ln (\frac{7}{5})

+1

十進法表記

5.94413 \dots

解答ステップ

\int_{0}^{1} 35 + 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{1}{35})} 35 sec^{3} (u) d u

定数を除く:

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

= 35 \cdot \int_{0}^{a r c t a n (\frac{1}{35})} sec^{3} (u) d u

積分換算を適用する

= 35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{1}{35})} sec (u) d u

\int_{0}^{a r c t a n (\frac{1}{35})} sec (u) d u = \frac{1}{2} ln (\frac{7}{5})

= 35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \frac{1}{2} ln (\frac{7}{5})

簡素化

35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \frac{1}{2} ln (\frac{7}{5}) : 35 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{35})} + \frac{35}{4} ln (\frac{7}{5})

= 35 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{35})} + \frac{35}{4} ln (\frac{7}{5})

限度を計算する:

\frac{3}{35}

= 35 \cdot \frac{3}{35} + \frac{35}{4} ln (\frac{7}{5})

簡素化

= 3 + \frac{35}{4} ln (\frac{7}{5})

グラフ

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