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人気のある微分積分 >

integral from 0 to 4 of 4sqrt(x^2+9)

解

\int_{0}^{4} 4 x^{2} + 9 d x

解

36 (\frac{10}{9} + \frac{1}{2} ln (3))

+1

十進法表記

59.77502 \dots

解答ステップ

\int_{0}^{4} 4 x^{2} + 9 d x

定数を除く:

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

= 4 \cdot \int_{0}^{4} x^{2} + 9 d x

三角関数による置換を適用する

= 4 \cdot \int_{0}^{a r c t a n (\frac{4}{3})} 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

= 4 \cdot 9 \cdot \int_{0}^{a r c t a n (\frac{4}{3})} sec^{3} (u) d u

積分換算を適用する

= 4 \cdot 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{4}{3})} sec (u) d u

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

= 4 \cdot 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3)

簡素化

4 \cdot 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3) : 36 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3))

= 36 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3))

限度を計算する:

\frac{10}{9}

= 36 (\frac{10}{9} + \frac{1}{2} ln (3))

グラフ

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