integral from 0 to 5 of e^{sqrt(3x+1)}

解答

\int_{0}^{5} e^{3 x + 1} d x

2 e^{4}

十进制

109.19630 \dots

求解步骤

\int_{0}^{5} e^{3 x + 1} d x

使用换元积分法

= \int_{1}^{4} \frac{2 e ^{u} u}{3} d u

提出常数:

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

= \frac{2}{3} \cdot \int_{1}^{4} e^{u} u d u

使用分布积分法

= \frac{2}{3} [e^{u} u - \int e^{u} d u]_{1}^{4}

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

= \frac{2}{3} [e^{u} u - e^{u}]_{1}^{4}

带入上下限计算后得:

3 e^{4}

= \frac{2}{3} \cdot 3 e^{4}

化简

= 2 e^{4}