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人気のある微分積分 >

integral from 0 to 4pi of sqrt(5+t^2)

解

\int_{0}^{4 π} 5 + t^{2} d t

解

2 π 16 π^{2} + 5 + \frac{5}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

+1

十進法表記

86.26528 \dots

解答ステップ

\int_{0}^{4 π} 5 + t^{2} d t

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{4 π}{5})} 5 sec^{3} (u) d u

定数を除く:

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

= 5 \cdot \int_{0}^{a r c t a n (\frac{4 π}{5})} sec^{3} (u) d u

積分換算を適用する

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4 π}{5})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{4 π}{5})} sec (u) d u

\int_{0}^{a r c t a n (\frac{4 π}{5})} sec (u) d u = ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4 π}{5})} + \frac{1}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

簡素化

5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4 π}{5})} + \frac{1}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1) : 5 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4 π}{5})} + \frac{5}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

= 5 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4 π}{5})} + \frac{5}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

限度を計算する:

\frac{2 π 16 π ^{2} + 5}{5}

= 5 \cdot \frac{2 π 16 π ^{2} + 5}{5} + \frac{5}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

簡素化

= 2 π 16 π^{2} + 5 + \frac{5}{2} ln (\frac{4 π}{5} + \frac{16 π ^{2}}{5} + 1)

グラフ

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