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人気のある微分積分 >

integral from 2 to 4 of sqrt(8+x^2)

解

\int_{2}^{4} 8 + x^{2} d x

解

8 (\frac{3}{2} - \frac{3}{4}) + 4 ln (\frac{2 3 + 3 2 - 2 - 6}{2})

+1

十進法表記

8.28480 \dots

解答ステップ

\int_{2}^{4} 8 + x^{2} d x

三角関数による置換を適用する

= \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} 8 sec^{3} (u) d u

定数を除く:

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

= 8 \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} sec^{3} (u) d u

積分換算を適用する

= 8 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} sec (u) d u

\int_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} sec (u) d u = ln (\frac{2 3 + 3 2 - 2 - 6}{2})

= 8 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} + \frac{1}{2} ln (\frac{2 3 + 3 2 - 2 - 6}{2})

簡素化

8 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} + \frac{1}{2} ln (\frac{2 3 + 3 2 - 2 - 6}{2}) : 8 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} + 4 ln (\frac{2 3 + 3 2 - 2 - 6}{2})

= 8 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (2)} + 4 ln (\frac{2 3 + 3 2 - 2 - 6}{2})

限度を計算する:

\frac{3}{2} - \frac{3}{4}

= 8 (\frac{3}{2} - \frac{3}{4}) + 4 ln (\frac{2 3 + 3 2 - 2 - 6}{2})

グラフ

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