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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

11.83154 \dots

解答ステップ

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

三角関数による置換を適用する

= \int_{0}^{a r c t a n (2)} 4 sec^{3} (u) d u

定数を除く:

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

= 4 \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

積分換算を適用する

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

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

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

簡素化

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

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

限度を計算する:

5

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

グラフ

人気の例

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