How Can I Solve This Improper Integral With Absolute Value?

[dan]

Hi there
I'm having trouble solving this integral!

Evaluate the improper integral
int[1/(sqrt{absolute value(1-4x^2)})]dx where the upper limit=1, lower limit=0

without ignoring the absolute value sign?


Thank you in advance

 

[HallsofIvy]

What exactly do you mean by "without ignoring the absolute value"?
Since this has an absolute value, one CAN'T do while ignoring the absolute value!

The obvious way to integrate this is to separate into two integrals, one from 0 to 1/2 and the other from 1/2 to 1:
integral (x=0 to 1/2) (1/sqrt(1- 4x^2))dx+ integral(x= 1/2 to 1)(1/sqrt(4x^2- 1))dx and then use trigonometric substitutions.

Using the definition of absolute value is certainly NOT ignoring it!

 

[tacman]

for any help!

Hi there! Improper integrals can definitely be tricky, but I'd be happy to help you with this one. In order to solve this integral, we will need to break it up into two parts and then use a substitution to simplify it. Let's take a look at the first part, from 0 to 1.

int[0 to 1]1/(sqrt{absolute value(1-4x^2)})dx

First, we can rewrite the absolute value as a piecewise function, where we consider the positive and negative cases separately.

1/(sqrt{1-4x^2}) for x>=0
1/(sqrt{4x^2-1}) for x<0

Next, we can make a substitution for the first part, using u = 1-4x^2, which gives us du = -8xdx and dx = -du/(8x). This will help us simplify the integral.

int[0 to 1]1/(sqrt{1-4x^2})dx = int[1 to 0] -1/(8x*sqrt{u})du
= -1/8 int[1 to 0] u^(-1/2)du
= -1/8[2u^(1/2)] from 1 to 0
= -1/4(1-2^(1/2))

Now, for the second part, we can make a similar substitution, using v = 4x^2 - 1, which gives us dv = 8xdx and dx = dv/(8x). This will give us:

int[0 to 1]1/(sqrt{4x^2-1})dx = int[-1 to 0] 1/(8x*sqrt{v})dv
= 1/8 int[-1 to 0] v^(-1/2)dv
= 1/8[-2v^(1/2)] from -1 to 0
= 1/4(2^(1/2)-1)

Now, we can add these two parts together to get the final answer:

int[0 to 1]1/(sqrt{absolute value(1-4x^2)})dx = -1/4(1-2^(1/2)) + 1/4(2^(1/2)-1