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受欢迎的微积分 >

integral from 0 to 2 of e^{x^2-2x}

解答

\int_{0}^{2} e^{x^{2} - 2 x} d x

解答

\frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

求解步骤

\int_{0}^{2} e^{x^{2} - 2 x} d x

对

x^{2} - 2 x

配方:

(x - 1)^{2} - 1

= \int_{0}^{2} e^{(x - 1)^{2} - 1} d x

使用换元积分法

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

化简

e^{u^{2} - 1} : e^{u^{2}} e^{- 1}

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

提出常数:

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

= e^{- 1} \cdot \int_{- 1}^{1} e^{u^{2}} d u

使用指数法则:

a^{- 1} = \frac{1}{a}

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

这是非初等函数积分:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= \frac{1}{e} [\frac{π}{2} erfi (u)]_{- 1}^{1}

带入上下限计算后得:

\frac{π erfi ( 1 ) - π erfi ( - 1 )}{2}

= \frac{1}{e} \cdot \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2}

化简

= \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

流行的例子

integral from 0 to 1 of (x)^2

\int_{0}^{1} (x)^{2} d x

integral from 1 to 3 of 2e^{-2x}

\int_{1}^{3} 2 e^{- 2 x} d x

integral from 2 to 5 of 6x

\int_{2}^{5} 6 x d x

integral from 1 to 5 of |2x-6|

\int_{1}^{5} ∣ 2 x - 6 ∣ d x

integral from 0 to L of x^2sin((npix)/L)

\int_{0}^{L} x^{2} sin (\frac{nπ x}{L}) d x