What is the integral from-2 to 1 of sqrt(3t^2-1) ?

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

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

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

diverges

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

\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)

Take the denominator(s) of

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

and compare to zero:

u = \frac{π}{2}

If exist

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

undefined

, \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 =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 2}^{2} 3 - ∣ x ∣^{3} d x

\int_{\frac{π}{2}}^{π} sec (x) tan (x) 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

What is the integral from-2 to 1 of sqrt(3t^2-1) ?
The integral from-2 to 1 of sqrt(3t^2-1) is diverges