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受欢迎的微积分 >

integral from 0 to 1 of sqrt(2+5x^2)

解答

\int_{0}^{1} 2 + 5 x^{2} d x

解答

\frac{1}{5} (\frac{35}{2} + ln (\frac{5}{2} + \frac{7}{2}))

+1

十进制

1.87694 \dots

求解步骤

\int_{0}^{1} 2 + 5 x^{2} d x

使用三角换元法

= \int_{0}^{a r c t a n (\frac{5}{2})} \frac{2}{5} sec^{3} (u) d u

提出常数:

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

= \frac{2}{5} \cdot \int_{0}^{a r c t a n (\frac{5}{2})} sec^{3} (u) d u

使用降次积分法

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

\int_{0}^{a r c t a n (\frac{5}{2})} sec (u) d u = ln (\frac{5}{2} + \frac{7}{2})

= \frac{2}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (\frac{5}{2} + \frac{7}{2})

化简

\frac{2}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (\frac{5}{2} + \frac{7}{2}) : \frac{1}{5} 2 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + ln (\frac{5}{2} + \frac{7}{2})

= \frac{1}{5} 2 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + ln (\frac{5}{2} + \frac{7}{2})

带入上下限计算后得:

\frac{35}{4}

= \frac{1}{5} (2 \cdot \frac{35}{4} + ln (\frac{5}{2} + \frac{7}{2}))

化简

= \frac{1}{5} (\frac{35}{2} + ln (\frac{5}{2} + \frac{7}{2}))

作图

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