integral of e^{-10x^2}

解答

\int e^{- 10 x^{2}} d x

\frac{π}{2 10} erf (10 x) + C

求解步骤

\int e^{- 10 x^{2}} d x

= \int e^{- (10 x)^{2}} d x

使用换元积分法

= \int e^{- u^{2}} \frac{1}{10} 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 e^{- u^{2}} d u

这是非初等函数积分:

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

= \frac{1}{10} \cdot \frac{π}{2} erf (u)

u = 10 x

代回

= \frac{1}{10} \cdot \frac{π}{2} erf (10 x)

化简

\frac{1}{10} \cdot \frac{π}{2} erf (10 x) : \frac{π}{2 10} erf (10 x)

= \frac{π}{2 10} erf (10 x)

解答补常数

= \frac{π}{2 10} erf (10 x) + C

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

y^{^{'}} + y = x y^{2}, y (0) = \frac{1}{2}

\int - \frac{sin ( ln ( x ) ) sin ( 2 ln ( x ) )}{2 x} d x

\int \frac{4 x ^{2} - 2 x + 1}{4 x ^{3} - x} d x

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