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 4 of sqrt(1+e^{-2y)}

解答

\int_{0}^{4} 1 + e^{- 2 y} d y

解答

- \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

+1

十进制

4.22590 \dots

求解步骤

\int_{0}^{4} 1 + e^{- 2 y} d y

使用换元积分法

= \int_{1}^{\frac{1}{e ^{8}}} - \frac{1 + u}{2 u} d u

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

= - \int_{\frac{1}{e ^{8}}}^{1} - \frac{1 + u}{2 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_{\frac{1}{e ^{8}}}^{1} \frac{1 + u}{u} d u)

使用换元积分法

= - (- \frac{1}{2} \cdot \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{2 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

= - (- \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{v ^{2}}{v ^{2} - 1} d v)

因式分解

v^{2} - 1 : - (- v^{2} + 1)

= - (- \frac{1}{2} \cdot 2 \cdot \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{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

= - (- \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{8}}}^{2} \frac{v ^{2}}{- v ^{2} + 1} d v))

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

= - (- \frac{1}{2} \cdot 2 (- \int_{1 + \frac{1}{e ^{8}}}^{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

= - (- \frac{1}{2} \cdot 2 (- (\int_{1 + \frac{1}{e ^{8}}}^{2} \frac{1}{- v ^{2} + 1} d v - \int_{1 + \frac{1}{e ^{8}}}^{2} 1 d v)))

\int_{1 + \frac{1}{e ^{8}}}^{2} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2}

\int_{1 + \frac{1}{e ^{8}}}^{2} 1 d v = 2 - 1 + \frac{1}{e ^{8}}

= - - \frac{1}{2} \cdot 2 - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} - (2 - 1 + \frac{1}{e ^{8}})

化简

- - \frac{1}{2} \cdot 2 - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} - (2 - 1 + \frac{1}{e ^{8}}) : - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

= - \frac{ln ( 2 + 1 ) - ln ( ( 2 - 1 ) ( 1 + \frac{1}{e ^{8}} + 1 ) ) + ln ( 1 + \frac{1}{e ^{8}} - 1 )}{2} + 2 - 1 + \frac{1}{e ^{8}}

作图

流行的例子

integral of (28s+28)/((s^2+1)(s-1)^3)

\int \frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

(\partial)/(\partial x)(3x+4y^2+3x^2y^3)

\frac{\partial}{\partial x} (3 x + 4 y^{2} + 3 x^{2} y^{3})

integral of x-2pi

\int x - 2 π d x

integral of (x) 3/4 x^2-4x+8

\int (x) \frac{3}{4} x^{2} - 4 x + 8 d x

derivative of (2x^3(4x^2))

\frac{d}{d x} ((2 x^{3}) (4 x^{2}))