What is the integral from 2 to 4 of sqrt(x^2-4) ?

The integral from 2 to 4 of sqrt(x^2-4) is 4(-1/2 ln(2+sqrt(3))+sqrt(3))

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

\int_{2}^{4} x^{2} - 4 d x

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

Decimal

4.29428 \dots

Solution steps

\int_{2}^{4} x^{2} - 4 d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} 4 tan^{2} (u) sec (u) d u

= \int_{0}^{\frac{π}{3}} 4 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

= 4 \cdot \int_{0}^{\frac{π}{3}} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 4 \cdot \int_{0}^{\frac{π}{3}} - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \int_{0}^{\frac{π}{3}} sec (u) d u + \int_{0}^{\frac{π}{3}} sec^{3} (u) d u)

\int_{0}^{\frac{π}{3}} sec (u) d u = ln (2 + 3)

\int_{0}^{\frac{π}{3}} sec^{3} (u) d u = 3 + \frac{1}{2} ln (2 + 3)

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

Simplify

- ln (2 + 3) + 3 + \frac{1}{2} ln (2 + 3) : - \frac{1}{2} ln (2 + 3) + 3

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

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What is the integral from 2 to 4 of sqrt(x^2-4) ?
The integral from 2 to 4 of sqrt(x^2-4) is 4(-1/2 ln(2+sqrt(3))+sqrt(3))