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人気のある微分積分 >

integral from 0 to sqrt(3 of)sqrt(x^2+9)

解

\int_{0}^{3} x^{2} + 9 d x

解

9 (\frac{1}{3} + \frac{1}{4} ln (3))

+1

十進法表記

5.47187 \dots

解答ステップ

\int_{0}^{3} x^{2} + 9 d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{6}} 9 sec^{3} (u) d u

定数を除く:

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

= 9 \cdot \int_{0}^{\frac{π}{6}} sec^{3} (u) d u

積分換算を適用する

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} sec (u) d u)

\int_{0}^{\frac{π}{6}} sec (u) d u = \frac{1}{2} ln (3)

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

簡素化

9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

限度を計算する:

\frac{1}{3}

= 9 (\frac{1}{3} + \frac{1}{4} ln (3))

グラフ

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