How do you test the improper integral #int (2x-1)^-3dx# from #[0,1/2]# and evaluate if possible?

Question

Paisley Johnson · Accepted Answer

The integral is divergent.

To calculate the improper integral, proceed as follows : First, calculate the indefinite integral #int_0^t(2x-1)^-3dx# #=[-1/4(2x-1)^-2]_0^t# #=[1/4(2x-1)^-2]_t^0# #=1/4-1/4(2t-1)^-2# Second, determine the limits as #t->1/2# #lim_(t->1/2)int_0^t(2x-1)^-3dx# #=lim_(t->1/2)(1/4-1/4(2t-1)^-2)# #=1/4-1/4(1/0)^2# #=-oo#

Axel Alvarez · Answer

undefined To test the improper integral $ \int_{0}^{1/2} (2x - 1)^{-3} \, dx $, first check if the function $ (2x - 1)^{-3} $ is continuous on the interval $[0, 1/2]$. Since the function is continuous except at $x = 1/2$, the integral is improper due to a discontinuity at $x = 1/2$.

To evaluate the improper integral, split it into two integrals: $ \int_{0}^{1/2} (2x - 1)^{-3} \, dx = \lim_{t \to 1/2^-} \int_{0}^{t} (2x - 1)^{-3} \, dx + \lim_{t \to 1/2^+} \int_{t}^{1/2} (2x - 1)^{-3} \, dx $.

Now, compute each limit separately. For the first limit, integrate $ (2x - 1)^{-3} $ from $0$ to $t$, then take the limit as $t$ approaches $1/2$ from the left. For the second limit, integrate $ (2x - 1)^{-3} $ from $t$ to $1/2$, then take the limit as $t$ approaches $1/2$ from the right.

After evaluating the integrals and limits, you'll have the value of the improper integral if it exists.