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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

2.95788 \dots

解答ステップ

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

簡素化

1 + (- 2 x)^{2} : 1 + 4 x^{2}

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

\int_{- a r c t a n (2)}^{a r c t a n (2)} sec (u) d u = ln (2 + 5) - 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 (2)} + \frac{1}{2} (ln (2 + 5) - 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 (2)} + \frac{1}{2} (ln (2 + 5) - 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 (2)} + \frac{1}{2} (ln (2 + 5) - 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 (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

限度を計算する:

25

= \frac{1}{2} (25 + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

簡素化

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

グラフ

人気の例

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\int_{a}^{b} 2 x^{2} d x

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integral from 0 to 2pi of 4

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