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人気のある微分積分 >

integral from 0 to 5 of sqrt(11+x^2)

解

\int_{0}^{5} 11 + x^{2} d x

解

15 + \frac{11}{4} ln (11)

+1

十進法表記

21.59421 \dots

解答ステップ

\int_{0}^{5} 11 + 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{5}{11})} 11 sec^{3} (u) d u

定数を除く:

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

= 11 \cdot \int_{0}^{a r c t a n (\frac{5}{11})} sec^{3} (u) d u

積分換算を適用する

= 11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{5}{11})} sec (u) d u

\int_{0}^{a r c t a n (\frac{5}{11})} sec (u) d u = \frac{1}{2} ln (11)

= 11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \frac{1}{2} ln (11)

簡素化

11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \frac{1}{2} ln (11) : 11 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{5}{11})} + \frac{11}{4} ln (11)

= 11 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{5}{11})} + \frac{11}{4} ln (11)

限度を計算する:

\frac{15}{11}

= 11 \cdot \frac{15}{11} + \frac{11}{4} ln (11)

簡素化

= 15 + \frac{11}{4} ln (11)

グラフ

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