The graph is a parabola:

The parabola is symmetric, so we may instead find the length of the curve from (0,0) to (7, 49/2) and double it:

A formula for arc length is:

`int_a^b sqrt(1+(dy/dx)^2) dx`

Thus we want to find:

`2 int _0 ^7 sqrt(1+(dy/dx)^2) dx`

`y=(1/2)x^2`

`dy/dx...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The graph is a parabola:

The parabola is symmetric, so we may instead find the length of the curve from (0,0) to (7, 49/2) and double it:

A formula for arc length is:

`int_a^b sqrt(1+(dy/dx)^2) dx`

Thus we want to find:

`2 int _0 ^7 sqrt(1+(dy/dx)^2) dx`

`y=(1/2)x^2`

`dy/dx = x`

`1+(dy/dx)^2 = 1+x^2`

So we want to find:

`2 int_0^7 sqrt(1+x^2) dx`

We ignore the constant, and the definite integral for now:

`int sqrt(1+x^2)dx`

We can make the trig substitution:

`x="tan" theta` , `dx = "sec"^2 theta d theta`

The integral becomes:

`int sqrt(1+"tan"^2 theta) "sec"^2 theta d theta`

`=int "sec"^3 theta d theta`

This is an integration by parts problem:

`u= "sec" theta` `dv="sec"^2 theta d theta`

`du = "sec" theta "tan" theta` `v= "tan" theta`

`int "sec"^3 theta d theta = "sec" theta "tan" theta - int "sec" theta "tan"^2 theta d theta`

`="sec" theta "tan" theta - int "sec" theta ("sec"^2 theta - 1) d theta`

`="sec" theta "tan" theta - int "sec"^3 theta d theta + int "sec" theta d theta`

`="sec" theta "tan" theta - int "sec" ^3 theta d theta + "ln" |"sec" theta + "tan" theta|`

Now, we have a `int "sec"^3 theta d theta` on both sides of the equation, so:

`2 int "sec"^3 theta d theta = "sec" theta "tan" theta + "ln" |"sec" theta + "tan" theta|`

`int "sec"^3 theta = (1/2) ( "sec" theta "tan" theta + "ln" |"sec" theta + "tan" theta| ) +C`

We have our answer in `theta` but we want it in x. Our original substitution was:

`x="tan" theta`

Another way to say this is `"tan" theta = x/1`

That is, if we had a right triangle, we could make `x` the opposite side of the angle, and 1 the adjacent side, and our triangle would represent this substitution.

Then the hypotenuse would be `sqrt(x^2+1)`

So `"sec" theta` would be `(sqrt(x^2+1))/1`

So:

`int sqrt(1+x^2) dx = (1/2)( (sqrt(x^2+1))(x) + "ln" |(sqrt(1+x^2))+(x)| )`

`=(1/2)(xsqrt(1+x^2) + "ln"|sqrt(1+x^2)+x|)+C`

Now we go back to our original problem:

`int_0^7 sqrt(1+x^2) = `

`(1/2)(7 sqrt(50)+"ln"|sqrt(50)+7|)-(1/2)(0sqrt(1) + "ln"(1+0))`

**`=(1/2)(7sqrt(50) + "ln" |sqrt(50)+7|)` **