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人気のある微分積分 >

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

解

\int_{1}^{2} 1 + (2 x)^{2} d x

解

17 - \frac{5}{2} + \frac{1}{4} ln (4 + 17) - \frac{1}{4} ln (2 + 5)

+1

十進法表記

3.16784 \dots

解答ステップ

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

積分換算を適用する

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

\int_{a r c t a n (2)}^{a r c t a n (4)} sec (u) d u = ln (4 + 17) - ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (2 + 5)))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (2 + 5)))

限度を計算する:

217 - 5

= \frac{1}{2} (217 - 5 + \frac{1}{2} (ln (4 + 17) - ln (2 + 5)))

簡素化

= 17 - \frac{5}{2} + \frac{1}{4} ln (4 + 17) - \frac{1}{4} ln (2 + 5)

グラフ

人気の例

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\int_{0}^{t} cos (wt) d t

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