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受欢迎的微积分 >

integral from 0 to 1 of 3te^{t^2-1}

解答

\int_{0}^{1} 3 t e^{t^{2} - 1} d t

解答

\frac{3 ( e - 1 )}{2 e}

+1

十进制

0.94818 \dots

求解步骤

\int_{0}^{1} 3 t e^{t^{2} - 1} d t

提出常数:

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

= 3 \cdot \int_{0}^{1} t e^{t^{2} - 1} d t

使用换元积分法

= 3 \cdot \int_{- 1}^{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

= 3 \cdot \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u

使用积分常见定则:

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

= 3 \cdot \frac{1}{2} [e^{u}]_{- 1}^{0}

化简

3 \cdot \frac{1}{2} [e^{u}]_{- 1}^{0} : \frac{3}{2} [e^{u}]_{- 1}^{0}

= \frac{3}{2} [e^{u}]_{- 1}^{0}

带入上下限计算后得:

1 - \frac{1}{e}

= \frac{3}{2} (1 - \frac{1}{e})

化简

= \frac{3 ( e - 1 )}{2 e}

作图

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