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受欢迎的微积分 >

integral from 1 to 2 of 21x^24^{x^3}

解答

\int_{1}^{2} 21 x^{2} 4^{x^{3}} d x

解答

\frac{229362}{ln ( 2 )}

+1

十进制

330899.41996 \dots

求解步骤

\int_{1}^{2} 21 x^{2} \cdot 4^{x^{3}} d x

提出常数:

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

= 21 \cdot \int_{1}^{2} x^{2} \cdot 4^{x^{3}} d x

使用换元积分法

= 21 \cdot \int_{4}^{65536} \frac{u}{3 ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

提出常数:

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

= 21 \cdot \frac{1}{3 ln ( 2 )} \cdot \int_{4}^{65536} \frac{u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} d u

使用分布积分法

= 21 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - \int - \frac{1}{4} d u]_{4}^{65536}

\int - \frac{1}{4} d u = - \frac{1}{4} u

= 21 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u)]_{4}^{65536}

化简

21 \cdot \frac{1}{3 ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} - (- \frac{1}{4} u)]_{4}^{65536} : \frac{7}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{4}^{65536}

= \frac{7}{ln ( 2 )} [\frac{u ^{2}}{2 ^{\frac{l n ( 2 u ) + l n ( 2 )}{l n ( 2 )}}} + \frac{1}{4} u]_{4}^{65536}

带入上下限计算后得:

32766

= \frac{7}{ln ( 2 )} \cdot 32766

化简

= \frac{229362}{ln ( 2 )}

作图

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\int 2 x^{2} y + 4 d y

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\int_{0}^{16} 2 x - \frac{x}{2} d x

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x \to - 1 lim (∣ x ∣)

y^{''}-4y^'+13y=0

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