integral from 0 to 0.6 of 10000x

解答

\int_{0}^{0.6} 10000 x d x

1800

求解步骤

\int_{0}^{0.6} 10000 x d x

提出常数:

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

= 10000 \cdot \int_{0}^{0.6} x d x

使用幂法则

= 10000 [\frac{x ^{2}}{2}]_{0}^{0.6}

化简

= 10000 [0.5 x^{2}]_{0}^{0.6}

带入上下限计算后得:

0.18

= 10000 \cdot 0.18

化简

= 1800

d er i v a t i v e 9 x^{3}

\frac{d}{d t} (\frac{3 t ^{2}}{1 + 4 t ^{2} + 9 t ^{4}})

\int e^{x^{2}} x^{3} d x

\int_{0}^{4} 2 π (2 - y) (4 y^{2}) d y

d er i v a t i v e \frac{e ^{x}}{e ^{x} + 1}