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人気のある微分積分 >

integral from 0 to 2 of sqrt(1+(2-2x)^2)

解

\int_{0}^{2} 1 + (2 - 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_{0}^{2} 1 + (2 - 2 x)^{2} d x

拡張

1 + (2 - 2 x)^{2} : 4 x^{2} - 8 x + 5

= \int_{0}^{2} 4 x^{2} - 8 x + 5 d x

平方完成する

4 x^{2} - 8 x + 5 : 4 (x - 1)^{2} + 1

= \int_{0}^{2} 4 (x - 1)^{2} + 1 d x

u置換積分法を適用する

= \int_{- 1}^{1} 4 u^{2} + 1 d u

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} (v) d v

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} (v) d v

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{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 (v) d v)

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

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{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} ( v ) sin ( v )}{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 (v) tan (v)]_{- 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 (v) tan (v)]_{- 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

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