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受欢迎的微积分 >

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

解答

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

解答

5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

+1

十进制

2.95788 \dots

求解步骤

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

使用换元积分法

= \int_{- 2}^{2} \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_{- 2}^{2} 1 + u^{2} d u

使用三角换元法

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

使用降次积分法

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

\int_{- a r c t a n (2)}^{a r c t a n (2)} sec (v) d v = ln (2 + 5) - ln (- 2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

化简

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

带入上下限计算后得:

25

= \frac{1}{2} (25 + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

化简

= 5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

作图

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