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受欢迎的微积分 >

integral from 0 to 5 of sqrt(11+x^2)

解答

\int_{0}^{5} 11 + x^{2} d x

解答

15 + \frac{11}{4} ln (11)

+1

十进制

21.59421 \dots

求解步骤

\int_{0}^{5} 11 + x^{2} d x

使用三角换元法

= \int_{0}^{a r c t a n (\frac{5}{11})} 11 sec^{3} (u) d u

提出常数:

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

= 11 \cdot \int_{0}^{a r c t a n (\frac{5}{11})} sec^{3} (u) d u

使用降次积分法

= 11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{5}{11})} sec (u) d u

\int_{0}^{a r c t a n (\frac{5}{11})} sec (u) d u = \frac{1}{2} ln (11)

= 11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \frac{1}{2} ln (11)

化简

11 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{11})} + \frac{1}{2} \cdot \frac{1}{2} ln (11) : 11 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{5}{11})} + \frac{11}{4} ln (11)

= 11 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{5}{11})} + \frac{11}{4} ln (11)

带入上下限计算后得:

\frac{15}{11}

= 11 \cdot \frac{15}{11} + \frac{11}{4} ln (11)

化简

= 15 + \frac{11}{4} ln (11)

作图

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