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 2 of sqrt(x^2+4)

解答

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

解答

2 (ln (1 + 2) + 2)

+1

十进制

4.59117 \dots

求解步骤

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

使用三角换元法

= \int_{0}^{\frac{π}{4}} 4 sec^{3} (u) d u

提出常数:

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

= 4 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

使用降次积分法

= 4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

化简

4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

带入上下限计算后得:

\frac{1}{2}

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

化简

= 2 (ln (1 + 2) + 2)

作图

流行的例子

integral from 0 to pi/6 of cos^6(3x)

integral from 1 to e of (ln(x))^9

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

integral from 2 to 3 of x/((x^2-2)^2)

integral from 2 to 6 of (1-t)^2