# 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 of the definite integral

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

there was a typo error. it must be from point (-2,5) to (4,2)

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)