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人気のある微分積分 >

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

解

\int_{0}^{π} 1 + x^{2} d x

解

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

+1

十進法表記

6.10991 \dots

解答ステップ

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

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

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

簡素化

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

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

限度を計算する:

\frac{π π ^{2} + 1}{2}

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

簡素化

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

グラフ

人気の例

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