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人気のある微分積分 >

integral from 0 to 3 of sqrt(1+4x^2)

解

\int_{0}^{3} 1 + 4 x^{2} d x

解

\frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

+1

十進法表記

9.74708 \dots

解答ステップ

\int_{0}^{3} 1 + 4 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 (6)} \frac{1}{2} sec^{3} (u) d u

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec (u) d u)

\int_{0}^{a r c t a n (6)} sec (u) d u = ln (6 + 37)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

限度を計算する:

337

= \frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

グラフ

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