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

解答

\int_{- 2}^{1} 3 t^{2} - 1 d t

发散

求解步骤

\int_{- 2}^{1} 3 t^{2} - 1 d t

使用三角换元法

= \int_{arcsec (- 23)}^{arcsec (3)} \frac{1}{3} tan^{2} (u) sec (u) d u

提出常数:

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

= \frac{1}{3} \cdot \int_{arcsec (- 23)}^{arcsec (3)} tan^{2} (u) sec (u) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{3} (- \int_{arcsec (3)}^{arcsec (- 23)} tan^{2} (u) sec (u) d u)

取

tan^{2} (u) sec (u)

的分母，令其等于零:

u = \frac{π}{2}

如果存在

b, a < b < c, f (b) =

无定义

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \frac{1}{3} (- (\int_{arcsec (3)}^{\frac{π}{2}} tan^{2} (u) sec (u) d u + \int_{\frac{π}{2}}^{arcsec (- 23)} tan^{2} (u) sec (u) d u))

\int_{arcsec (3)}^{\frac{π}{2}} tan^{2} (u) sec (u) d u =

发散

如果表达式有部分发散，则整个表达式发散

= 发散

\int_{- 2}^{2} 3 - ∣ x ∣^{3} d x

\int_{0}^{y} \frac{1}{1 - x} d x

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

\int_{0}^{2 π} 4 sin (t) d t

\int_{- 1}^{0} 3^{x - 1} d x