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人気のある微分積分 >

integral from 0 to 4 of sqrt(1+4x^2)

解

\int_{0}^{4} 1 + 4 x^{2} d x

解

\frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

+1

十進法表記

16.81863 \dots

解答ステップ

\int_{0}^{4} 1 + 4 x^{2} d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (8)} \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_{0}^{a r c t a n (8)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec (u) d u)

\int_{0}^{a r c t a n (8)} sec (u) d u = ln (8 + 65)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

限度を計算する:

465

= \frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

グラフ

人気の例

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

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

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\int_{0}^{a} e^{- x} d x

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\int_{- 12}^{6} (- 2 x + 12) d x

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