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受欢迎的微积分 >

integral from 0 to 1 of 2e^{x/3+4}

解答

\int_{0}^{1} 2 e^{\frac{x}{3} + 4} d x

解答

6 e^{4} (3 e - 1)

+1

十进制

129.59823 \dots

求解步骤

\int_{0}^{1} 2 e^{\frac{x}{3} + 4} d x

提出常数:

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

= 2 \cdot \int_{0}^{1} e^{\frac{x}{3} + 4} d x

e^{\frac{x}{3} + 4} = e^{4} e^{\frac{x}{3}}

= 2 \cdot \int_{0}^{1} e^{4} e^{\frac{x}{3}} d x

提出常数:

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

= 2 e^{4} \cdot \int_{0}^{1} e^{\frac{x}{3}} d x

使用换元积分法

= 2 e^{4} \cdot \int_{0}^{\frac{1}{3}} e^{u} \cdot 3 d u

提出常数:

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

= 2 e^{4} \cdot 3 \cdot \int_{0}^{\frac{1}{3}} e^{u} d u

使用积分常见定则:

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

= 2 e^{4} \cdot 3 [e^{u}]_{0}^{\frac{1}{3}}

化简

= 6 e^{4} [e^{u}]_{0}^{\frac{1}{3}}

带入上下限计算后得:

3 e - 1

= 6 e^{4} (3 e - 1)

作图

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