# Thread: Help with a simple Integration

1. ## Help with a simple Integration

Hello ,
I need to calculate the following line length , any help is appreciated
please tell me the way and not only the final result , thanks

calculate the following line length (0 =< X =< X0) , y^2 = 2px

i got the following integral :

int[x=0 to x=x0] Sqrt(1+p/(2x))dx

i dunno how to go from there...

Hello ,
I need to calculate the following line length , any help is appreciated
please tell me the way and not only the final result , thanks

calculate the following line length (0 =< X =< X0) , y^2 = 2px

i got the following integral :

int[x=0 to x=x0] Sqrt(1+p/(2x))dx

i dunno how to go from there...
The graph of the relation y^2 = 2px (I assume p is some constant) is not a function; there are 2 y-values for every x-value in this graph's domain except where x = 0. Since it's not a function, calculating the "line length" from 0 to x0 means we'll be integrating along 2 separate paths (the upper and lower path of this sideways parabola). If I'm mistaken, please correct me, but if I'm understanding this question correctly, I don't think enough information has been given to solve this.

Also, by "line length" do you mean "arc-length," as in 'the length of the curve of the graph from 0 to x0'?

One last question. In the integration you got, int[x=0 to x=x0] Sqrt(1+p/(2x))dx, where did you get p/(2x)?

3. ## Explanation:

You are right , i took the positive value of the root and thought to double the answer that i will get as its symetric.

i used the following formula :
if we have y=f(x) and we want to calculate the length of the curve as you pointed from 0 to x0
then it is int[from x=0 to x=x0] [Sqrt(1+(f(x))^2)]dx

we have y^2 = 2px
y = +/- sqrt(2px) , i took +sqrt(2px)
y = (1/2sqrt(2px))2p = p / sqrt(2px)

(y)^2 = p^2/(2px) = p/(2x)

so we need to calculate : int[0 to x0] Sqrt(1+p/(2x)) dx

i hope i am not mistaken... anyway from here i just couldnt use any substitue since i cant change the values when x=0 ..
if you have any ideas...

You are right , i took the positive value of the root and thought to double the answer that i will get as its symetric.

i used the following formula :
if we have y=f(x) and we want to calculate the length of the curve as you pointed from 0 to x0
then it is int[from x=0 to x=x0] [Sqrt(1+(f(x))^2)]dx

we have y^2 = 2px
y = +/- sqrt(2px) , i took +sqrt(2px)
y = (1/2sqrt(2px))2p = p / sqrt(2px)

(y)^2 = p^2/(2px) = p/(2x)

so we need to calculate : int[0 to x0] Sqrt(1+p/(2x)) dx

i hope i am not mistaken... anyway from here i just couldnt use any substitue since i cant change the values when x=0 ..
if you have any ideas...
Thank you for the clarification. Everything you did is correct. I just wanted to be sure I understood the question you were asking.

I have a suggestion. Don't integrate this function horizontally. Try this instead:

INT[-y(x0) to +y(x0)] Sqrt(1 + (x')^2)dx

To do this, we need the vertical limits of this function.
Since it's a horizontal parabola, we will be integrating from y = -sqrt(2px0) to y = +sqrt(2px0)
Now, to find x'
x = y^2/(2p) --> x' = 2y/(2p) = y/p

Now, integrate:
INT[-sqrt(2px0) to sqrt(2px0)] Sqrt(1 + (y/p)^2)

Integrating this will involve a double substitution, but it is possible. Do you know how to use trig substitution?

5. ## Thanks

Yes i am familiar with that .

Thanks for the idea , i will try to think it over and let you know if i have any problems .

6. Originally Posted by ecMathGeek
Thank you for the clarification. Everything you did is correct. I just wanted to be sure I understood the question you were asking.

I have a suggestion. Don't integrate this function horizontally. Try this instead:

INT[-y(x0) to +y(x0)] Sqrt(1 + (x')^2)dx

To do this, we need the vertical limits of this function.
Since it's a horizontal parabola, we will be integrating from y = -sqrt(2px0) to y = +sqrt(2px0)
Now, to find x'
x = y^2/(2p) --> x' = 2y/(2p) = y/p

Now, integrate:
INT[-sqrt(2px0) to sqrt(2px0)] Sqrt(1 + (y/p)^2) dy

Integrating this will involve a double substitution, but it is possible. Do you know how to use trig substitution?
I'll show you what to do. Hopefully this makes sense:

Let y/p = tanu <-- this is similar to 'u' substitution except we're using a function of u instead of just u
dy/p = (secu)^2 du --> dy = p(secu)^2 du

For now, I'm going to ignore the limits of integration:

INT [Sqrt(1 + (tanu)^2)*p(secu)^2] du ... Now, for the reason I used tanu in this substitution, I know that 1 + (tanu)^2 = (secu)^2
p*INT [Sqrt((secu)^2)(secu)^2] du ... the Sqrt((secu)^2) = secu
p*INT (secu)^3 du
p*INT 1/(cosu)^3 du
p*INT 1/[(cosu)^2*cosu] du ... now I'll use the identity (cosu)^2 = 1 - (sinu)^2
p*INT 1/[(1 - (sinu)^2)*cosu] du

Here's where the second substitution comes in:
Let v = sinu
dv = cosu du --> du = dv/cosu

p*INT 1/[(1 - v^2)*(cosu)] dv/cosu
p*INT 1/[(1 - v^2)*(cosu)^2)] dv ... now, let's use (cosu)^2 = 1 - (sinu)^2, where sinu = v
p*INT 1/[(1 - v^2)*(1 - v^2)] dv

Wow ... lol. This problem is more complicated than I thought. (To any helper in these forums) Can someone check to see if I'm even on the right track. I know this can be solved from here, and I know how to solve it, but it's turning out to be a lot more complicated that I anticipated.

If not, I'll pick up where I left off in a few minutes. (Hopefully I have helped you some)