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人気のある微分積分 >

integral from 0 to 4 of sqrt(1+(x/3)^2)

解

\int_{0}^{4} 1 + (\frac{x}{3})^{2} d x

解

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

+1

十進法表記

4.98125 \dots

解答ステップ

\int_{0}^{4} 1 + (\frac{x}{3})^{2} d x

簡素化

1 + (\frac{x}{3})^{2} : \frac{1}{3} 9 + x^{2}

= \int_{0}^{4} \frac{1}{3} 9 + x^{2} d x

定数を除く:

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

= \frac{1}{3} \cdot \int_{0}^{4} 9 + x^{2} d x

三角関数による置換を適用する

= \frac{1}{3} \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

= \frac{1}{3} \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

= \frac{1}{3} \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)

= \frac{1}{3} \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

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

= 3 ([\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}

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

グラフ

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