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受欢迎的微积分 >

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

解答

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

解答

3 + \frac{1}{2 2} ln (22 + 3)

+1

十进制

3.62322 \dots

求解步骤

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

使用三角换元法

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

\int_{0}^{a r c t a n (2 \cdot 2)} sec (u) d u = ln (22 + 3)

= \frac{1}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 \cdot 2)} + \frac{1}{2} ln (22 + 3)

化简

\frac{1}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 \cdot 2)} + \frac{1}{2} ln (22 + 3) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (22)} + \frac{1}{2} ln (22 + 3))

带入上下限计算后得:

32

= \frac{1}{2} (32 + \frac{1}{2} ln (22 + 3))

化简

= 3 + \frac{1}{2 2} ln (22 + 3)

作图

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