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人気のある微分積分 >

integral from 0 to 2 of sqrt(1+3x^2)

解

\int_{0}^{2} 1 + 3 x^{2} d x

解

\frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

+1

十進法表記

4.17014 \dots

解答ステップ

\int_{0}^{2} 1 + 3 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 (3 \cdot 2)} \frac{1}{3} 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}{3} \cdot \int_{0}^{a r c t a n (3 \cdot 2)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (3 \cdot 2)} sec (u) d u

\int_{0}^{a r c t a n (3 \cdot 2)} sec (u) d u = ln (23 + 13)

= \frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13)

簡素化

\frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13) : \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

= \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

限度を計算する:

39

= \frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

グラフ

人気の例

tangent 6x^{-3},\at x=5

t an g e n t 6 x^{- 3}, at x = 5

derivative of (x^2/(sqrt(x)))

\frac{d}{d x} (\frac{x ^{2}}{x})

integral from 0 to 1 of |2x^2-1|

\int_{0}^{1} 2 x^{2} - 1 d x

(dx)/(dt)=2x(5-x)

\frac{d x}{d t} = 2 x (5 - x)

y^{^{'}} = e^{2 y + 6 x}