What is the integral from 0 to 2 of sqrt(1+(2y-2)^2) ?

The integral from 0 to 2 of sqrt(1+(2y-2)^2) is sqrt(5)+1/4 ln(2+sqrt(5))-1/4 ln(-2+sqrt(5))

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integral from 0 to 2 of sqrt(1+(2y-2)^2)

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

Decimal Notation

2.95788 \dots

Solution steps

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

Apply u-substitution

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

= \int_{- 2}^{2} \frac{1 + u ^{2}}{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_{- 2}^{2} 1 + u^{2} d u

Apply Trigonometric Substitution

: \int_{- a r c t a n (2)}^{a r c t a n (2)} sec^{3} (v) d v

= \frac{1}{2} \cdot \int_{- a r c t a n (2)}^{a r c t a n (2)} sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{- a r c t a n (2)}^{a r c t a n (2)} sec (v) d v)

\int_{- a r c t a n (2)}^{a r c t a n (2)} sec (v) d v = ln (2 + 5) - ln (- 2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Compute the boundaries:

25

= \frac{1}{2} (25 + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Simplify

= 5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

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What is the integral from 0 to 2 of sqrt(1+(2y-2)^2) ?
The integral from 0 to 2 of sqrt(1+(2y-2)^2) is sqrt(5)+1/4 ln(2+sqrt(5))-1/4 ln(-2+sqrt(5))