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人気のある微分積分 >

integral from 0 to pi/4 of sqrt(x^2+1)

解

\int_{0}^{\frac{π}{4}} x^{2} + 1 d x

解

\frac{π π ^{2} + 16}{32} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

+1

十進法表記

0.85995 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} x^{2} + 1 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{π}{4})} 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 (\frac{π}{4})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{π}{4})} sec (u) d u

\int_{0}^{a r c t a n (\frac{π}{4})} sec (u) d u = ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

限度を計算する:

\frac{π π ^{2} + 16}{32}

= \frac{π π ^{2} + 16}{32} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

グラフ

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