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人気のある微分積分 >

integral from 0 to 5 of sqrt(1+x^2)

解

\int_{0}^{5} 1 + x^{2} d x

解

\frac{5 13}{2} + \frac{1}{2} ln (5 + 26)

+1

十進法表記

13.90376 \dots

解答ステップ

\int_{0}^{5} 1 + 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 (5)} sec^{3} (u) d u

積分換算を適用する

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

\int_{0}^{a r c t a n (5)} sec (u) d u = ln (5 + 26)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5)} + \frac{1}{2} ln (5 + 26)

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5)} + \frac{1}{2} ln (5 + 26) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (5)} + \frac{1}{2} ln (5 + 26)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (5)} + \frac{1}{2} ln (5 + 26)

限度を計算する:

\frac{5 13}{2}

= \frac{5 13}{2} + \frac{1}{2} ln (5 + 26)

グラフ

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