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人気のある微分積分 >

integral from 0 to 1 of sqrt(1+8t^2)

解

\int_{0}^{1} 1 + 8 t^{2} d t

解

\frac{3}{2} + \frac{1}{4 2} ln (22 + 3)

+1

十進法表記

1.81161 \dots

解答ステップ

\int_{0}^{1} 1 + 8 t^{2} d t

三角関数による置換を適用する

= \int_{0}^{a r c t a n (22)} \frac{1}{2 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 2} \cdot \int_{0}^{a r c t a n (22)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (22)} sec (u) d u

\int_{0}^{a r c t a n (22)} sec (u) d u = ln (22 + 3)

= \frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3)

簡素化

\frac{1}{2 2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3) : \frac{1}{2 2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

= \frac{1}{2 2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

限度を計算する:

32

= \frac{1}{2 2} (32 + \frac{1}{2} ln (22 + 3))

簡素化

= \frac{3}{2} + \frac{1}{4 2} ln (22 + 3)

グラフ

人気の例

integral from 0 to 1 of pi(y-y^2)^2

\int_{0}^{1} π (y - y^{2})^{2} d y

integral from 2 to \sqrt[3]{3 of} 1/x

\int_{2}^{33} \frac{1}{x} d x

integral from x to 2 of t

\int_{x}^{2} t d t

integral from 0 to x/2 of sqrt(sin(x))

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

integral from 1 to infinity of 8/(x^5)

\int_{1}^{\infty} \frac{8}{x ^{5}} d x