integral from 0 to 2 of sqrt((1+3x^2))

解答

\int_{0}^{2} (1 + 3 x^{2}) d x

\frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

十进制

4.17014 \dots

求解步骤

\int_{0}^{2} 1 + 3 x^{2} d x

使用三角换元法

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

\int_{0}^{a r c t a n (3 \cdot 2)} sec (u) d u = ln (23 + 13)

= \frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13)

化简

\frac{1}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3 \cdot 2)} + \frac{1}{2} ln (23 + 13) : \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

= \frac{1}{3} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (23)} + \frac{1}{2} ln (23 + 13))

带入上下限计算后得:

39

= \frac{1}{3} (39 + \frac{1}{2} ln (23 + 13))

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