# Math Help - Evaluate sig x(x+1)^1/2 .dx

1. ## Evaluate sig x(x+1)^1/2 .dx

Could someone give me a leg-up for this question?

2. ## Re: Evaluate sig x(x+1)^1/2 .dx

Hey Darrylcwc.

What does sig mean? Is it the signum function? Something else?

3. ## Re: Evaluate sig x(x+1)^1/2 .dx

Yes it refers to the sigma function

4. ## Re: Evaluate sig x(x+1)^1/2 .dx

Is the range restricted to x >= -1? (In order to have only real solutions)?

If so then you should consider the intervals [-1,0) and [0,whatever]. Also you will have to put a constraint on the limits if you want a finite answer.

Recall that x(x+1)^(1/2) is an increasing function which should make your job easier.

5. ## Re: Evaluate sig x(x+1)^1/2 .dx

Sorry I wasn't clear. I need to find the integral.

6. ## Re: Evaluate sig x(x+1)^1/2 .dx

Basically the signum function is -1 if x < 0, 0 if x = 0 and 1 if x > 0.

You need to find when the function is < 0, 0, and > 0.

x(x+1)^(1/2) = 0 implies x = 0 or x = -1.

This means sig (x(x+1)^1/2) when x > 0 = 1 and this means Integral 0 to u sig(x(x+1)^(1/2)) = Integral 0 to u 1dx = u - 0 = u.

Now you need to find when x(x+1)^(1/2) < 0 and for that region use sig(x(x+1)^(1/2)) = -1 when the condition is satisfied.

7. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by Darrylcwc
Sorry I wasn't clear. I need to find the integral.
Try change of variable such as u = x + 1, and rewrite the sqrt as a power 1/2

8. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by votan
Try change of variable such as u = x + 1, and rewrite the sqrt as a power 1/2
Did so before you suggested. It was futile.

The range is from 2 to -2: The integral I arrived at was: [u^2/2 - 10u].2u^(0.5); it's wrong.

9. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by Darrylcwc
Did so before you suggested. It was futile.

The range is from 2 to -2: The integral I arrived at was: [u^2/2 - 10u].2u^(0.5); it's wrong.

are your limits of integration on u or on x. If they are for x convert them to u and you are set

10. ## Re: Evaluate sig x(x+1)^1/2 .dx

Forget that way: just find when the expression inside the signum function is positive, negative, and 0 and replace it with +1, -1, and 0 respectively and integrate it piece-wise. (For example if u is positive for sig(u) then replace sig(u(x))dx with +1dx.)

11. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by votan

are your limits of integration on u or on x. If they are for x convert them to u and you are set
What about integrate x/[(3x+10)^0.5].dx from the range of 2 to -2
I let u = 3x + 10 and x = [u-10]/3
du/dx = 3
dx = du/3
I keep arriving at [2u^3/2]/3 - 20u^(0.5)

Tried but again not working out

12. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by Darrylcwc
What about integrate x/[(3x+10)^0.5].dx from the range of 2 to -2
I let u = 3x + 10 and x = [u-10]/3
du/dx = 3
dx = du/3
I keep arriving at [2u^3/2]/3 - 20u^(0.5)

Tried but again not working out
Are you integrating y = x(3x+10)^(-0.5) or signum function sig[y]

13. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by votan
Are you integrating y = x(3x+10)^(-0.5) or signum function sig[y]
I'm trying to integrate the function x(3x+10)^(-0.5) and thereafter taking the function f(2) - f(-2)

The antiderivative I arrived at is [2u^3/2]/3 - 20u^(0.5)

When I tried f(2) - f(-2), i keep getting the wrong answer.

14. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by Darrylcwc
I'm trying to integrate the function x(3x+10)^(-0.5) and thereafter taking the function f(2) - f(-2)

The antiderivative I arrived at is [2u^3/2]/3 - 20u^(0.5)

When I tried f(2) - f(-2), i keep getting the wrong answer.

15. ## Re: Evaluate sig x(x+1)^1/2 .dx

Originally Posted by votan
Solved.