The arc of the parabola y=1/4(x-2)^2 + 1 from the point (2,5) to the point (4,2).length of arc of the graph of the function additional applications...

Length = integral(sqrt(1+(y')^2) dx, a,
b)

y' = 1/4 * 2(x-2) =
1/2(x-2)

(y')^2 =
1/4(x-2)^2

1+(y')^2 = 1/4(x-2)^2 + 1 = 1/4(x^2 - 4x + 4) +
1 = 1/4(x^2 - 4x + 8)

so Length =
integral(1/2sqrt(x^2-4x+8), -2, 4)

use t = (x-2) +
sqrt((x-2)^2 + 4)) and substitute

1/2 integral(sqrt(x^2 -
4x + 8)
= ln (x - 2 + sqrt(x^2-4x+8)) - (1/2)sqrt(x²-4x+8) + (1/4) x
sqrt(x²-4x+8)

evaluate at x = 4 we get
= ln( 4 -
2+sqrt(16 - 16 + 8))-1/2 sqrt(16 - 16 + 8) + 1/4 (4)(sqrt(16-16+8)
= ln(2 +
sqrt(8)) - 1/2 sqrt(8) + sqrt(8))
= ln(2 + sqrt(8)) + 1/2
sqrt(8))
= ln(2 + 2sqrt(2)) + sqrt(2)

evaluate at
x = -2 we get

= ln( -4 + sqrt(4 + 8 + 8)) - 1/2 sqrt(4 + 8
+ 8) + 1/4(-2)sqrt(4 + 8 + 8)
= ln(-4 + sqrt(20)) - 1/2 sqrt(20) + -1/2
sqrt(20)
= ln(-4 + 2sqrt(5)) - sqrt(20)
= ln(-4 + 2sqrt(5)) - 2
sqrt(5)

So our answer is
ln(2+2sqrt(2)) +
sqrt(2) - ln(-4 + 2sqrt(5)) + 2sqrt(5)
= ln(2sqrt(2)+2) - ln(2sqrt(5) - 4) +
sqrt(2) + 2sqrt(5)