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受欢迎的微积分 >

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

解答

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

解答

\frac{e ^{4} - 1}{2 e ^{4}}

+1

十进制

0.49084 \dots

求解步骤

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

使用换元积分法

= \int_{0}^{- 4} - \frac{e ^{u}}{2} d u

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

= - \int_{- 4}^{0} - \frac{e ^{u}}{2} 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_{- 4}^{0} e^{u} d u)

使用积分常见定则:

\int e^{u} d u = e^{u}

= - (- \frac{1}{2} [e^{u}]_{- 4}^{0})

化简

= \frac{1}{2} [e^{u}]_{- 4}^{0}

带入上下限计算后得:

1 - \frac{1}{e ^{4}}

= \frac{1}{2} (1 - \frac{1}{e ^{4}})

化简

= \frac{e ^{4} - 1}{2 e ^{4}}

作图

流行的例子

integral from 0 to 2 of rsqrt(4-r^2)

\int_{0}^{2} r 4 - r^{2} d r

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

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

integral from 0 to y of 2e^{-(x+y)}

\int_{0}^{y} 2 e^{- (x + y)} d x

integral from-1 to 5 of (3x^4+2x^2+3)

\int_{- 1}^{5} (3 x^{4} + 2 x^{2} + 3) d x

integral from 0 to 5 of-3x

\int_{0}^{5} - 3 x d x