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受欢迎的微积分 >

integral from 0 to 4 of 2*6^{-2x}

解答

\int_{0}^{4} 2 \cdot 6^{- 2 x} d x

解答

\frac{1679615}{1679616 ln ( 6 )}

+1

十进制

0.55811 \dots

求解步骤

\int_{0}^{4} 2 \cdot 6^{- 2 x} d x

提出常数:

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

= 2 \cdot \int_{0}^{4} 6^{- 2 x} d x

使用换元积分法

= 2 \cdot \int_{0}^{- 8} - 3^{u} \cdot 2^{u - 1} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 8}^{0} - 3^{u} \cdot 2^{u - 1} d u)

提出常数:

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

= 2 (- (- \int_{- 8}^{0} 3^{u} \cdot 2^{u - 1} d u))

化简

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= 2 (- (- \int_{- 8}^{0} 3^{u} \cdot 2^{u} \cdot 2^{- 1} d u))

提出常数:

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

= 2 (- (- 2^{- 1} \cdot \int_{- 8}^{0} 3^{u} \cdot 2^{u} d u))

使用指数法则:

a^{- 1} = \frac{1}{a}

= 2 (- (- \frac{1}{2} \cdot \int_{- 8}^{0} 3^{u} \cdot 2^{u} d u))

化简

3^{u} \cdot 2^{u} : 6^{u}

= 2 (- (- \frac{1}{2} \cdot \int_{- 8}^{0} 6^{u} d u))

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 2 (- (- \frac{1}{2} [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}))

化简

2 (- (- \frac{1}{2} [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0})) : [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}

= [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}

带入上下限计算后得:

\frac{1679615}{1679616 ln ( 6 )}

= \frac{1679615}{1679616 ln ( 6 )}

作图

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