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

解答

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

\frac{1}{2} (\frac{3}{2} + \frac{1}{2} ln (2 + 3))

十进制

1.27127 \dots

求解步骤

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

使用三角换元法

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

使用降次积分法

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

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

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

化简

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

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

带入上下限计算后得:

\frac{3}{2}

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