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人気のある微分積分 >

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

解

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

解

\frac{1}{2} (417 + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

+1

十進法表記

9.29356 \dots

解答ステップ

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

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

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

簡素化

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

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

限度を計算する:

417

= \frac{1}{2} (417 + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

グラフ

人気の例

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\int_{3}^{9} (\frac{1}{x ^{2}}) d x

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\int_{4}^{16} x - 1 d x

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