Evaluate the definite integral

Question

Evaluate the definite integral:

∫ Lower Limit is 0; upper limit is 2 and the function is 1/(2x-3)^2 dx

I've been doing an assignment where I am evaluating definite integrals and I've been finding the anti-derivative and substituting U for these problems, but I am extremely lost because when I evaluated this problem by substituting U I got -2/3, but that was wrong, and I ended up checking what the answer was and it was, "does not exist." I am hoping someone can explain this problem in a video so I can understand how to determine whether a definite integral exists or not.

Answer 1

Answer 2

integral from 0 to 2 of 1/(2x-3)²dx. You can let u = 2x-3 and du = 2dx (x = (1/2) du) That would be the normal procedure, but the integrand blows up at x = 3/2, so you have to split the integral up into two pieces from 0 to 3/2 and 3/2 to 2 and that both must be finite. This is not true for either integral. The integral is the area under the curve, f, only if f is bounded in the interval.

In order to text the integral of the an unbounded function, you have to take the limit as n goes to 3/2 of the integral from 0 to n. You evaluate the integral and find that the -1/(2(2n-3)) integral term goes to infinity as n approaches 3/2.