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-2 to 2 of sqrt(1+100x^2)

解答

\int_{- 2}^{2} 1 + 100 x^{2} d x

解答

\frac{1}{10} (20401 + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

+1

十进制

40.41891 \dots

求解步骤

\int_{- 2}^{2} 1 + 100 x^{2} d x

使用三角换元法

= \int_{- a r c t a n (20)}^{a r c t a n (20)} \frac{1}{10} 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}{10} \cdot \int_{- a r c t a n (20)}^{a r c t a n (20)} sec^{3} (u) d u

使用降次积分法

= \frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} \cdot \int_{- a r c t a n (20)}^{a r c t a n (20)} sec (u) d u)

\int_{- a r c t a n (20)}^{a r c t a n (20)} sec (u) d u = ln (20 + 401) - ln (- 20 + 401)

= \frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

化简

\frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401))) : \frac{1}{10} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

= \frac{1}{10} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

带入上下限计算后得:

20401

= \frac{1}{10} (20401 + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

作图

流行的例子

f (x) = - cot (x)

integral of (3x^3+6x^2-12x+5)

\int (3 x^{3} + 6 x^{2} - 12 x + 5) d x

derivative f(x)=t^5sin(t)

d er i v a t i v e f (x) = t^{5} sin (t)

integral from 0 to 1 of sqrt(1+144x)

\int_{0}^{1} 1 + 144 x d x

derivative t^2sqrt(t)

d er i v a t i v e t^{2} t