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人気のある微分積分 >

integral from 0 to 3 of sqrt(1+1296x^2)

解

\int_{0}^{3} 1 + 1296 x^{2} d x

解

\frac{1}{36} (5411665 + \frac{1}{2} ln (108 + 11665))

+1

十進法表記

162.08160 \dots

解答ステップ

\int_{0}^{3} 1 + 1296 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 (108)} \frac{1}{36} 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}{36} \cdot \int_{0}^{a r c t a n (108)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (108)} sec (u) d u)

\int_{0}^{a r c t a n (108)} sec (u) d u = ln (108 + 11665)

= \frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

簡素化

\frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665)) : \frac{1}{36} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

= \frac{1}{36} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

限度を計算する:

5411665

= \frac{1}{36} (5411665 + \frac{1}{2} ln (108 + 11665))

グラフ

人気の例

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