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受欢迎的微积分 >

integral from 0 to 2 of sqrt(1+36x^2)

解答

\int_{0}^{2} 1 + 36 x^{2} d x

解答

145 + \frac{1}{12} ln (12 + 145)

+1

十进制

12.30657 \dots

求解步骤

\int_{0}^{2} 1 + 36 x^{2} d x

使用三角换元法

= \int_{0}^{a r c t a n (12)} \frac{1}{6} sec^{3} (u) d u

提出常数:

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

= \frac{1}{6} \cdot \int_{0}^{a r c t a n (12)} sec^{3} (u) d u

使用降次积分法

= \frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (12)} sec (u) d u)

\int_{0}^{a r c t a n (12)} sec (u) d u = ln (12 + 145)

= \frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

化简

\frac{1}{6} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145)) : \frac{1}{6} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

= \frac{1}{6} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

带入上下限计算后得:

6145

= \frac{1}{6} (6145 + \frac{1}{2} ln (12 + 145))

化简

= 145 + \frac{1}{12} ln (12 + 145)

作图

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