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人気のある微分積分 >

integral from 0 to 5 of sqrt(x^2+25)

解

\int_{0}^{5} x^{2} + 25 d x

解

\frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

+1

十進法表記

28.69483 \dots

解答ステップ

\int_{0}^{5} x^{2} + 25 d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{4}} 25 sec^{3} (u) d u

定数を除く:

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

= 25 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

積分換算を適用する

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

簡素化

25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

限度を計算する:

\frac{1}{2}

= 25 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

簡素化

= \frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

グラフ

人気の例

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