integral from 0 to 1 of sqrt(θ^2+1)

解答

\int_{0}^{1} θ^{2} + 1 d θ

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

十进制

1.14779 \dots

求解步骤

\int_{0}^{1} θ^{2} + 1 d θ

使用三角换元法

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

使用降次积分法

= [\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)

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

化简

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

= [\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}

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

化简

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

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