00:01
For this problem, we need to determine whether the improper integral converges or diverges.
00:05
Now, this integrand is discontinuous at x equals 2, and so we rewrite this as the limit, as t approaches to from the right, of the integral from t to 4 of 2 over x times the square root of x squared minus 4 dx.
00:25
And then from here we apply trigonometric substitution.
00:27
We let x equal to 2 secan theta, and we get dx equal to 2 sikin theta, hem's tangent theta, d theta.
00:38
And so from here we have the limit, as t approaches 2 from the right, of the integral from t to 4 of, we have 2 over 2 sicken theta times the square root of the square root of the square of 2 secan theta, which is 4 sqin squared theta, minus 4 times d x which is 2 secantheta times tangent theta d theta now from here we can cancel out 2 secan theta and you're left with the limit as the approaches 2 from the right of the integral from t to 4 of 2 tangent theta over the square root of 4 times secan squared minus 1 d theta and this secan squared minus 1 is just tangent squared theta and so from here we simplify and we get limit st...