zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 0 to 3 of sqrt(1+4x^2)

解答

\int_{0}^{3} 1 + 4 x^{2} d x

解答

\frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

+1

十进制

9.74708 \dots

求解步骤

\int_{0}^{3} 1 + 4 x^{2} d x

使用三角换元法

= \int_{0}^{a r c t a n (6)} \frac{1}{2} sec^{3} (u) d u

提出常数:

\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 (6)} sec^{3} (u) d u

使用降次积分法

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec (u) d u)

\int_{0}^{a r c t a n (6)} sec (u) d u = ln (6 + 37)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

化简

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

带入上下限计算后得:

337

= \frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

作图

流行的例子

d/(dt)(at^2+bt+c)

\frac{d}{d t} (a t^{2} + b t + c)

d er i v a t i v e 2

integral of xsec^4(x^2+1)tan^7(x^2+1)

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

limit as x approaches 3 of 5/(x(x^3-27))

x \to 3 lim (\frac{5}{x ( x ^{3} - 27 )})

integral of (8sqrt(x))/(x^3+1)

\int \frac{8 x}{x ^{3} + 1} d x