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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

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

derivative y=3+5sqrt(x)

d er i v a t i v e y = 3 + 5 x

(\partial)/(\partial x)(cosh(x))

\frac{\partial}{\partial x} (cosh (x))

integral of (5r^2)/(r+1)

\int \frac{5 r ^{2}}{r + 1} d r

tangent f(x)=9x^2+5

t an g e n t f (x) = 9 x^{2} + 5

sum from n=0 to infinity of 1/(n^{3/2)}

n = 0 \sum \infty \frac{1}{n ^{\frac{3}{2}}}