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受欢迎的微积分 >

integral from-1 to 3 of sqrt(1+(2y+2)^2)

解答

\int_{- 1}^{3} 1 + (2 y + 2)^{2} d y

解答

\frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

+1

十进制

16.81863 \dots

求解步骤

\int_{- 1}^{3} 1 + (2 y + 2)^{2} d y

使用换元积分法

= \int_{0}^{8} \frac{1 + u ^{2}}{2} 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}^{8} 1 + u^{2} d u

使用三角换元法

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec^{3} (v) d v

使用降次积分法

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec (v) d v)

\int_{0}^{a r c t a n (8)} sec (v) d v = ln (8 + 65)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

化简

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65)) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

带入上下限计算后得:

465

= \frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

作图

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