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受欢迎的微积分 >

integral from 0 to 1 of sqrt(1+9x^2)

解答

\int_{0}^{1} 1 + 9 x^{2} d x

解答

\frac{5}{2} + \frac{1}{6} ln (3 + 10)

+1

十进制

1.88421 \dots

求解步骤

\int_{0}^{1} 1 + 9 x^{2} d x

使用三角换元法

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

使用降次积分法

= \frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (3)} sec (u) d u)

\int_{0}^{a r c t a n (3)} sec (u) d u = ln (3 + 10)

= \frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

化简

\frac{1}{3} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)) : \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

= \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10))

带入上下限计算后得:

\frac{3 5}{2}

= \frac{1}{3} (\frac{3 5}{2} + \frac{1}{2} ln (3 + 10))

化简

= \frac{5}{2} + \frac{1}{6} ln (3 + 10)

作图

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