`(1,2) , (7,10)` Find the distance between the two points using integration.

 Given the equation of a line `y = mx + b,`

=> slope = `dy/dx = m` . Thus, the distance is:

`L = int_a^b sqrt(1+(dy/dx)^2) dx` where `a<=x<=b`

we know the two points `(x_1,y_1)=(1,2)`

`(x_2,y_2)=(7,10)`

`m = (y_2- y_1)/(x_2-x_1) = (10-2)/(7-1) = 8/6=4/3`

so now the length is` L = int_1^7 sqrt(1+(4/3)^2) dx`

 =` int_1^7 sqrt(1+(16/9)) dx`

 =` int_1^7 sqrt(25/9) dx`

= `int_1^7 (5/3) dx`

= `(5/3) int_1^7 1 dx` 

= `(5/3) |_1^7 x`

...

 Given the equation of a line `y = mx + b,`

=> slope = `dy/dx = m` . Thus, the distance is:

`L = int_a^b sqrt(1+(dy/dx)^2) dx` where `a<=x<=b`

we know the two points `(x_1,y_1)=(1,2)`

`(x_2,y_2)=(7,10)`

`m = (y_2- y_1)/(x_2-x_1) = (10-2)/(7-1) = 8/6=4/3`

so now the length is` L = int_1^7 sqrt(1+(4/3)^2) dx`

 =` int_1^7 sqrt(1+(16/9)) dx`

 =` int_1^7 sqrt(25/9) dx`

= `int_1^7 (5/3) dx`

= `(5/3) int_1^7 1 dx` 

= `(5/3) |_1^7 x`

=` (5/3)[7-1]`

=` (5/3)6 = 5*2 = 10`

so the distance between the two points = 10