integral from 0 to t of xe^{x^2}

解答

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

\frac{1}{2} (e^{t^{2}} - 1)

求解步骤

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

使用换元积分法

= \int_{0}^{t^{2}} \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_{0}^{t^{2}} e^{u} d u

使用积分常见定则:

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

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

带入上下限计算后得:

e^{t^{2}} - 1

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

\int_{0}^{3} t e^{- s t} d t

\int_{0}^{9} (\frac{10}{( 1 + x ) ^{2}}) d x

\int_{0}^{2} - x^{2} + x + 1 d x

\int_{4}^{9} 2 t (10 t + 3) d t

\int_{0}^{2} 3 x - x^{2} d x