Find the average distance between (5, 6) to a point on the line y = x (5 < x < 10)

Note that the coordinates of any point on the line can be written as:
(x, x).

Then, the distance between (x, x) and (5, 6) is:
D(x) = √[(x - 5)^2 + (x - 6)^2] = √(2x^2 - 22x + 61).

We wish to find the average value of D(x) on the interval [5, 10]. Using the formula for the average value of a function, this is:
D(avg) = 1/(10 - 5) ∫ √(2x^2 - 22x + 61) dx (from x=5 to 10)
= 1/5 ∫ √(2x^2 - 22x + 61) dx (from x=5 to 10).

If we complete the square under the radical, we will get:
2x^2 - 22x + 61 = 2(x^2 - 11x) + 61
= 2(x - 11/2)^2 + 61 - 2(121/4)
= 2(x - 11/2)^2 + 1/2.

Then:
D(avg) = 1/5 ∫ √(2x^2 - 22x + 61) dx (from x=5 to 10)
= 1/5 ∫ √[2(x - 11/2)^2 + 1/2] dx (from x=5 to 10)
= √2/5 ∫ √[(x - 11/2)^2 + (1/2)^2] dx (from x=5 to 10), by pulling out 2 from the radical.

To evaluate ∫ √[(x - 11/2)^2 + (1/2)^2] dx, we need to apply the following substitution:
(1/2)tanθ = x - 11/2 ==> x = (tanθ + 11)/2, dx = (1/2)sec^2θ, tanθ = 2x - 11.

Then, changing the bounds:
(a) Lower bound: x = 5
==> tanθ = 2(5) - 11 = -1, θ = 3π/4, and secθ = √[(-1)^2 + 1] = √2.
(b) Upper bound: x = 10
==> tanθ = 2(10) - 11 = 9, θ = arctan(9), and secθ = √(1 + 9^2) = √82.
(You will see why I calculated the values of secθ when I apply the limits.)

This gives:
D(avg) = √2/5 ∫ √[(x - 11/2)^2 + (1/2)^2] dx (from x=5 to 10)
= √2/10 ∫ sec^2θ√[(1/4)tan^2θ + 1/4] dθ (from θ=3π/4 to arctan(9))
= √2/20 ∫ sec^3θ dθ (from θ=3π/4 to arctan(9)), since tan^2θ + 1 = sec^2θ
= (√2/40)(secθtanθ + ln|secθ + tanθ|) (evaluated from θ=3π/4 to arctan(9))
= (√2/40){[(√82)(9) + ln|√82 + 9|] - [(√2)(-1) + ln|√2 - 1|]}
(I plugged in the values of secθ and tanθ here.)
= (√2/40)[9√82 + ln(9 + √82) + √2 - ln(√2 - 1)] ≈ 3.06.

I hope this helps!