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受欢迎的微积分 >

integral from-1 to 3 of sqrt(1+4x^2)

解答

\int_{- 1}^{3} 1 + 4 x^{2} d x

解答

\frac{3 37 + 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (- 2 + 5)

+1

十进制

11.22603 \dots

求解步骤

\int_{- 1}^{3} 1 + 4 x^{2} d x

使用三角换元法

= \int_{- a r c t a n (2)}^{a r c t a n (6)} \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_{- a r c t a n (2)}^{a r c t a n (6)} sec^{3} (u) d u

使用降次积分法

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

\int_{- a r c t a n (2)}^{a r c t a n (6)} sec (u) d u = ln (6 + 37) - ln (- 2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (- 2 + 5)))

化简

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (- 2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (- 2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (- 2 + 5)))

带入上下限计算后得:

337 + 5

= \frac{1}{2} (337 + 5 + \frac{1}{2} (ln (6 + 37) - ln (- 2 + 5)))

化简

= \frac{3 37 + 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (- 2 + 5)

作图

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