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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

1.47894 \dots

解答ステップ

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

簡素化

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

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

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

限度を計算する:

5

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

グラフ

人気の例

tangent f(x)=x^2+5,(3,14)

t an g e n t f (x) = x^{2} + 5, (3, 14)

derivative x/4-4/x

d er i v a t i v e \frac{x}{4} - \frac{4}{x}

x^2(dy)/(dx)-2xy=6y^4,y(1)= 1/3

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{3}

limit as x approaches infinity of 7x

x \to \infty lim (7 x)

integral of 1/(x+7)

\int \frac{1}{x + 7} d x