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受欢迎的微积分 >

integral from 0 to 5 of sqrt(25+x^2)

解答

\int_{0}^{5} 25 + x^{2} d x

解答

\frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

+1

十进制

28.69483 \dots

求解步骤

\int_{0}^{5} 25 + x^{2} d x

使用三角换元法

= \int_{0}^{\frac{π}{4}} 25 sec^{3} (u) d u

提出常数:

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

= 25 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

使用降次积分法

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

化简

25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

带入上下限计算后得:

\frac{1}{2}

= 25 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

化简

= \frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

作图

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