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 1 to 5 of sqrt(x^2+4)

解

\int_{1}^{5} x^{2} + 4 d x

解

4 (\frac{5 29 - 5}{8} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

+1

十進法表記

14.67691 \dots

解答ステップ

\int_{1}^{5} x^{2} + 4 d x

三角関数による置換を適用する

= \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} 4 sec^{3} (u) d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec^{3} (u) d u

積分換算を適用する

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec (u) d u

\int_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} sec (u) d u = ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4})

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4})

簡素化

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}) : 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{1}{2})}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

限度を計算する:

\frac{5 29 - 5}{8}

= 4 (\frac{5 29 - 5}{8} + \frac{1}{2} ln (- \frac{( 5 + 29 ) ( 1 - 5 )}{4}))

グラフ

人気の例

integral from 0 to cos(x) of ysin(x)

\int_{0}^{c o s (x)} y sin (x) d y

integral from 0 to 7 of (2xsqrt(x^2-1))

\int_{0}^{7} (2 x x^{2} - 1) d x

integral from pi/3 to pi/6 of tan^2(x)

\int_{\frac{π}{3}}^{\frac{π}{6}} tan^{2} (x) d x

integral from-1 to 3 of e^x-3

\int_{- 1}^{3} e^{x} - 3 d x

integral from 0 to 1 of (1-e^2)^{1/2}

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