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人気のある微分積分 >

integral from 1 to 3 of sqrt(x^2+8)

解

\int_{1}^{3} x^{2} + 8 d x

解

\frac{3 17 - 3}{2} + ln (\frac{( 3 2 + 34 ) ^{4}}{1024})

+1

十進法表記

6.99285 \dots

解答ステップ

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

\int_{a r c t a n (\frac{1}{2 2})}^{a r c t a n (\frac{3}{2 2})} sec (u) d u = ln (\frac{3 2 + 34}{4}) - \frac{1}{2} ln (2)

= 8 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2 2})}^{a r c t a n (\frac{3}{2 2})} + \frac{1}{2} (ln (\frac{3 2 + 34}{4}) - \frac{1}{2} ln (2))

簡素化

8 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2 2})}^{a r c t a n (\frac{3}{2 2})} + \frac{1}{2} (ln (\frac{3 2 + 34}{4}) - \frac{1}{2} ln (2)) : 8 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{2}{4})}^{a r c t a n (\frac{3 2}{4})} + 4 ln (\frac{3 2 + 34}{4}) - 2 ln (2)

= 8 [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{2}{4})}^{a r c t a n (\frac{3 2}{4})} + 4 ln (\frac{3 2 + 34}{4}) - 2 ln (2)

限度を計算する:

\frac{3 17}{16} - \frac{3}{16}

= 8 (\frac{3 17}{16} - \frac{3}{16}) + 4 ln (\frac{3 2 + 34}{4}) - 2 ln (2)

簡素化

= \frac{3 17 - 3}{2} + ln (\frac{( 3 2 + 34 ) ^{4}}{1024})

グラフ

人気の例

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