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

The integral from 1 to 2 of sqrt(x^2-x) is (-ln(2sqrt(2)+3)+6sqrt(2))/8

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

\int_{1}^{2} x^{2} - x d x

\frac{- ln ( 2 2 + 3 ) + 6 2}{8}

Decimal Notation

0.84031 \dots

Solution steps

\int_{1}^{2} x^{2} - x d x

Complete the square

x^{2} - x : (x - \frac{1}{2})^{2} - \frac{1}{4}

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

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4 u ^{2} - 1}{2} d u

= \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4 u ^{2} - 1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} 4 u^{2} - 1 d u

Apply Trigonometric Substitution

: \int_{0}^{arcsec (3)} \frac{1}{2} tan^{2} (v) sec (v) d v

= \frac{1}{2} \cdot \int_{0}^{arcsec (3)} \frac{1}{2} tan^{2} (v) sec (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (3)} tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (3)} (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (3)} - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

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

= \frac{1}{2} \cdot \frac{1}{2} (- \int_{0}^{arcsec (3)} sec (v) d v + \int_{0}^{arcsec (3)} sec^{3} (v) d v)

\int_{0}^{arcsec (3)} sec (v) d v = ln (22 + 3)

\int_{0}^{arcsec (3)} sec^{3} (v) d v = 32 + \frac{1}{2} ln (22 + 3)

= \frac{1}{2} \cdot \frac{1}{2} (- ln (22 + 3) + 32 + \frac{1}{2} ln (22 + 3))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- ln (22 + 3) + 32 + \frac{1}{2} ln (22 + 3)) : \frac{- ln ( 2 2 + 3 ) + 6 2}{8}

= \frac{- ln ( 2 2 + 3 ) + 6 2}{8}

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What is the integral from 1 to 2 of sqrt(x^2-x) ?
The integral from 1 to 2 of sqrt(x^2-x) is (-ln(2sqrt(2)+3)+6sqrt(2))/8