ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to 2 of sqrt(1+5x^2)

解

\int_{0}^{2} 1 + 5 x^{2} d x

解

\frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

+1

十進法表記

5.07524 \dots

解答ステップ

\int_{0}^{2} 1 + 5 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 (5 \cdot 2)} \frac{1}{5} 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}{5} \cdot \int_{0}^{a r c t a n (5 \cdot 2)} sec^{3} (u) d u

積分換算を適用する

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

\int_{0}^{a r c t a n (5 \cdot 2)} sec (u) d u = ln (25 + 21)

= \frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21)

簡素化

\frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21) : \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

= \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

限度を計算する:

105

= \frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

グラフ

人気の例

integral from 7 to 9 of (x-7)^6

\int_{7}^{9} (x - 7)^{6} d x

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

\int_{2}^{3} \frac{2 x}{( 1 + x ^{2} )} d x

integral from 1 to 2 of sin(pix)

\int_{1}^{2} sin (π x) d x

integral from 0 to pi/2 of cos(x+2y)

\int_{0}^{\frac{π}{2}} cos (x + 2 y) d y

integral from 0 to 2 of (3-x)(x-4)

\int_{0}^{2} (3 - x) (x - 4) d x