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受欢迎的微积分 >

integral from-1 to 3 of sqrt(1+(6y+2)^2)

解答

\int_{- 1}^{3} 1 + (6 y + 2)^{2} d y

解答

\frac{5 401 + 17}{3} + \frac{1}{12} ln (20 + 401) - \frac{1}{12} ln (- 4 + 17)

+1

十进制

35.23136 \dots

求解步骤

\int_{- 1}^{3} 1 + (6 y + 2)^{2} d y

使用换元积分法

= \int_{- 4}^{20} \frac{1 + u ^{2}}{6} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{- 4}^{20} 1 + u^{2} d u

使用三角换元法

= \frac{1}{6} \cdot \int_{- a r c t a n (4)}^{a r c t a n (20)} sec^{3} (v) d v

使用降次积分法

= \frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (20)} sec (v) d v)

\int_{- a r c t a n (4)}^{a r c t a n (20)} sec (v) d v = ln (20 + 401) - ln (- 4 + 17)

= \frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

化简

\frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17))) : \frac{1}{6} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

= \frac{1}{6} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

带入上下限计算后得:

10401 + 217

= \frac{1}{6} (10401 + 217 + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

化简

= \frac{5 401 + 17}{3} + \frac{1}{12} ln (20 + 401) - \frac{1}{12} ln (- 4 + 17)

作图

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