What is the integral from 12 to 16 of (y^2-6)^{1/2} ?

The integral from 12 to 16 of (y^2-6)^{1/2} is

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integral from 12 to 16 of (y^2-6)^{1/2}

\int_{12}^{16} (y^{2} - 6)^{\frac{1}{2}} d y

6 (- \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}})

Decimal Notation

55.13000 \dots

Solution steps

\int_{12}^{16} (y^{2} - 6)^{\frac{1}{2}} d y

Apply Trigonometric Substitution

: \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} 6 tan^{2} (u) sec (u) d u

= \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} 6 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

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} - 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

= 6 - \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec (u) d u + \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec^{3} (u) d u

\int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec (u) d u = ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})

\int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec^{3} (u) d u = \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}} + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}}))

= 6 (- (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}} + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})))

Add similar elements:

- (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}})) + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}})) = - \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}}))

= 6 (- \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}})

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What is the integral from 12 to 16 of (y^2-6)^{1/2} ?
The integral from 12 to 16 of (y^2-6)^{1/2} is