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 4 of (sqrt(16-x^2))/x

解答

\int_{2}^{4} \frac{16 - x ^{2}}{x} d x

解答

2 (ln (\frac{3}{2} + 1) - 3 - ln (- \frac{3}{2} + 1))

+1

十进制

1.80372 \dots

求解步骤

\int_{2}^{4} \frac{16 - x ^{2}}{x} d x

使用换元积分法

= \int_{23}^{0} - \frac{u ^{2}}{- u ^{2} + 16} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{23} - \frac{u ^{2}}{- u ^{2} + 16} d u

提出常数:

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

= - (- \int_{0}^{23} \frac{u ^{2}}{- u ^{2} + 16} d u)

使用换元积分法

= - (- \int_{0}^{\frac{3}{2}} \frac{4 v ^{2}}{- v ^{2} + 1} d v)

提出常数:

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

= - (- 4 \cdot \int_{0}^{\frac{3}{2}} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - (- 4 \cdot \int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} - 1 d v)

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- 4 (\int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} d v - \int_{0}^{\frac{3}{2}} 1 d v))

\int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} d v = \frac{1}{2} (ln (\frac{3}{2} + 1) - ln (- \frac{3}{2} + 1))

\int_{0}^{\frac{3}{2}} 1 d v = \frac{3}{2}

= - (- 4 (\frac{1}{2} (ln (\frac{3}{2} + 1) - ln (- \frac{3}{2} + 1)) - \frac{3}{2}))

化简

- (- 4 (\frac{1}{2} (ln (\frac{3}{2} + 1) - ln (- \frac{3}{2} + 1)) - \frac{3}{2})) : 2 (ln (\frac{3}{2} + 1) - 3 - ln (- \frac{3}{2} + 1))

= 2 (ln (\frac{3}{2} + 1) - 3 - ln (- \frac{3}{2} + 1))

作图

流行的例子

integral from x to 1 of 6(1-y)

\int_{x}^{1} 6 (1 - y) d y

integral from 0 to 2 of x^2+2x

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

integral from 1/e to e of ln(x)

\int_{\frac{1}{e}}^{e} ln (x) d x

integral from 0 to 5 of (x^3-1)

\int_{0}^{5} (x^{3} - 1) d x

integral from 1 to infinity of 4/(x^3)

\int_{1}^{\infty} \frac{4}{x ^{3}} d x